Considering the definition of molar mass, there are 162 grams of Al in 6 moles of Al.
Definition of molar massThe molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
Mass of 6 moles of AlYou want to determine the mass in 6 moles of Al. So, being the molar mass of Al is 27 g/mole, you can apply the following rule of three: If by definition of molar mass 1 mole of Al contains 27 grams, 6 moles of Al contains how much mass?
mass= (6 moles× 27 grams) ÷1 mole
mass= 162 grams
In summary, there are 162 grams of Al in 6 moles of Al.
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Which of the following is a physical change?
A) Magnesium reduces dry ice, forming black
carbon and MgO.
B) Water forms hydrogen and oxygen gases
via electrolysis.
C) The pH of water is raised by dissolving
sodium hydroxide.
D) Ice forms as water is poured onto dry ice.
E) Water hydrates cobalt chloride, turning
it from blue to pink.
Ice forms as water is poured onto dry ice is a physical change .
So , option D is correct one .
The physical change is explained in following points ,
A physical change is one that simply affects the physical nature of the substance without changing their chemical properties .Its chemical properties remain unchanged .Usually, increasing the temperature or applying pressure—or both—will cause a physical change .Reversing the condition—lowering the temperature, the pressure, or both—restores the material's initial state .In other words, physical changes are reversible .to learn more about physical change please click here ,
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How many total atoms are in 0.160 g of [tex]P_{2}[/tex][tex]O_{5}[/tex]?
total atoms:
The total number of atoms in 0.160g of P2O5 is 3 x 10^21 atoms.
What are atoms?
Atoms are defined as a smallest unit into which a matter can be divided without the release of electrically charged particles.
First calculate the number of mole of P2O5
Mass of P2O5 = 0.160 g
Molar mass of P2O5 = 142g/mole
Number of mole of P2O5 = mass/ molar mass
n = 0.160/142 g/mole
n = 0.0011 mole of P2O5
One mole of P2O5 has five mole of oxygen.
Using conversion factor
= 0.0011 mole of P2O5 x [ 5 mole of oxygen/ 1 mole of P2O5]
= 0.0055 mole of oxygen
Total number of atoms = 0.0055 x 6.02 x 10^23 atom/mole
= 3 x 10^21 atoms
Thus, the total number of atoms in 0.160g of P2O5 is 3 x 10^21 atoms.
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Is atomic radius a periodic property of atoms?
Major periodic trends include atomic radius, ionization energy, electron affinity, electronegativity, valency and metallic character.
Specify the hybridization at the atoms labelled a-d.
For each atom enter one of the following: sp3, dsp2, sp2, sp3d, or sp.
The orbitals where you can find each of the bonds;
a - sp3
b - sp3
c - sp2
d- sp3
What is hybridization?The term hybridization is the term that is used to describe the mixture of atomic orbitals which can be used to combine with other orbitals of relevant energy.
Let us recall that in some cases, the energy that is required in bonding could only be obtained by the hybridization of orbitals in which orbitals are mixed. The mixed orbitals are orbitals that are very close to each other in energy and help in the formation of bond.
Now let us try to write the orbitals where you can find each of the bonds;
a - sp3
b - sp3
c - sp2
d- sp3
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A metal (M) forms a compound with the formula MCl2 . The compound contains 57.69% Cl by mass. Identify the metal in the compound? Give the symbol of the metal.
A metal (M) forms a compound with formula MCl₂. The compound contains 57.69% Cl by mass and the metal is Chromium. Symbol of Chromium metal is Cr.
Given :
57.69 % Cl by mass means that 57.69 grams of chlorine present in 100 grams of compound.
Mass of compound = 100 g
Mass of chlorine = 57.69 g
Mass of metal = 100 - 57.69= 42.31 g
First, we will calculate the moles of chlorine, using formula
Moles = Given mass
molar mass
Moles = 57.69
35.45
Moles = 1.627 moles
The given formula of compound is, MCl₂
Therefore, ratio of M : Cl = 1 : 2
We will calculate the moles of metal,
2 moles of Cl combines with 1 mole of metal
1.627 mole of Cl combines with 1.627 = 0.813 mole of metal
2
calculate the molar mass of metal -
Molar mass of metal = Given mass
moles
Molar mass = 42.31 g = 52.04 g/mol
0.813 mol
The molar mass of metal is 52.04 g/mole.
From this, we can conclude that the metal is Chromium (Cr)
So, the formula will be, CrCl₂
Hence, the identity of the metal is Chromium (Cr)
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To tolerate biting forces that average 28,000 psi modern dental restorative materials have strengths in excess of
To tolerate biting forces that average 28,000 psi, modern dental restorative materials have strengths in excess of dna psi.
What is dna ?Deoxyribonucleic acid is a polymer made of two polynucleotide chains that coil around one another to form a double helix and which contains the genetic material necessary for the growth, development, and reproduction of all known organisms as well as a large number of viruses. The two types of nucleic acids are DNA and RNA. One of the four major types of macromolecules that are necessary for all known forms of life is nucleic acids, along with proteins, lipids, and complex carbohydrates (polysaccharides).
Due to the fact that the two DNA strands are made up of more straightforward monomeric units known as nucleotides, they are referred to as polynucleotides. Each nucleotide consists of a phosphate group, a sugar called deoxyribose, and one of the four nucleobases that contain nitrogen: cytosine (C), guanine (G), adenine (A), or thymine (T).
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(C3H8) What mass of carbon dioxide (in kg ) is produced upon the complete combustion of 40.2 L of propane (approximate contents of two 5-gallon tanks)? Assume that the density of the liquid propane in the tank is 0.621 g/mL . (Hint: Begin by writing a balanced equation for the combustion reaction.)
The mass of carbon dioxide is obtained as 236 g of carbon dioxide.
What mass of carbon dioxide is produced?We know that the first step when we are trying to solve any type of question that involves stoichiometry is that we have to obtain the balanced reaction equation. This would show us the kind of interaction that is able to take place in order to obtain the results that we seek.
The reaction equation is given as;
C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O
If 1 mole of propane occupies 22.4 L
x mole of propane occupies 40.2 L
x = 1 mole * 40.2 L /22.4 L
x = 1.79 moles
Now;
1 mole of propane produced 3 moles of carbon dioxide
1.79 moles of propane produced 1.79 moles * 3 moles/1 mole
= 5.37 moles of carbon dioxide
Mass of carbon dioxide =5.37 moles * 44 g/mol = 236 g of carbon dioxide.
The mass of carbon dioxide is obtained as 236 g of carbon dioxide.
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an isotopes of caesium-137 has a half-life of 30 years. if 1.0g of Cs-137 disintegrates over a period of 90 years, how many grams of Cs-137 would remain?
After 90 years amount of Cs-137 isotopes has a half-life of 30 years. would remain is equal to = 0.125 grams.
An important and useful factor of radioactive decay is half of life, which is the amount of time it takes for one half of a radioactive isotope to decay. The half of-life of a selected radioactive isotope is constant; it is unaffected by way of situations and is independent of the initial quantity of that isotope.
The half of life of a chemical reaction can be defined because the time is taken for the awareness of a given reactant to attain 50% of its preliminary awareness. The half of-life of a response is the time required for the reactant's attention to decrease to at least one-half of its initial price. The half of-life of a first-order response is consistent this is associated with the charge consistent for the reaction. t1/2 = 0.693/k
Since all radioactive reaction follows first-order reaction, The rate of reaction does not depend on initial concentration.
calculation:-
1 gram Cs-137 at T = 0
after the first half,
⇒0.5 gram Cs-137 at T = 30
After the second half,
⇒0.25 gram Cs-137 at T = 30.
After the third half,
⇒0.125 gram Cs-137 at T = 90 years.
Hence, after 90 years amount of Cs-137 would remain is equal to = 0.125 grams.
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Describe one way in which energy is transferred when magnesium combusts
Explanation:
When magnesium reacts with oxygen, it produces light bright enough to blind you temporarily. Magnesium burns so bright because the reaction releases a lot of heat. As a result of this exothermic reaction, magnesium gives two electrons to oxygen, forming powdery magnesium oxide (MgO).
Question 2 (3 points)
What is the empirical formula for a compound that contain 1.011 g of chromium and 0.467 g of oxygen. ?
Chromium oxide's empirical formula is Cr₂O₃
The simplest whole-number ratio of atoms in a compound is the empirical formula. The proportion of atoms to moles is the same. Our task then is to determine the molar ratio of Cr to O.
Cr's mass is 7.35 g.
10.74 g of chromium oxide are equal to 10.74 g of chromium and 10.74 g of oxygen.
O's mass is 10.74 g - 7.35 g, or 3.39 g.
Moles of Mg = 52.00 g Mg = 0.1413 mol Mg/7.35 g Mg = 1 mol Mg
Moles of O equal 3.39 g O to 1 mol O and 16.00 g O to 0.2119 mol O
For every 3 mol of O, there are 2 mol of Cr.
Chromium oxide's empirical formula is Cr₂O₃.
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what's the structure of benzene ?
anyone willing to talk rn ?
Answer:
The structure has a six-carbon ring which is represented by a hexagon and it includes 3-double bonds.
Karl sane walked 2189 miles in 45 days, 7 hours, and 39 minutes.
What was his average speed in miles per day?
What volume of silver will weigh exactly 2500g. The density of silver is 10g/cm3
Answer:
250 cm ^3
Explanation:
Do you consider comparing consumer products on the basis of their components for use, safety, quality, and cost?
Comparing consumer products on the basis of their components like use, safety, quality, and cost is necessary as it creates awareness about the product among consumers.
Consumer Products - It is a kind of product that is bought by individuals for personal use or consumption. These are the end products or final goods reached to the end-user i.e consumer by retailer via various distribution channels. Ex- grocery, clothing, etc.
Marketing strategies are built based on different market segmentation and types of consumer products available to the target market.
Types of Consumer Products
There are four types of consumer products, they are:-
Convenience Products
Shopping Products
Specialty Products
Unsought Products.
Hence, it is necessary to compare consumer products on basis of the components like use, safety, quality, and cost. It also increases the competition as well as creates awareness among consumers.
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Complete combustion of 8.90 g of a hydrocarbon produced 27.3 g of CO₂ and 13.0 g of H₂O. What is the empirical formula for
the hydrocarbon? Insert subscripts as necessary.
Empirical formula is C4H7.
Empirical formula- A formula that lists the relative amounts of the constituent components in a compound without specifying the number or configuration of atoms.
When a hydrocarbon (CxHy) burns, CO2 and H2O are produced.
(Note: It mentions that a hydrocarbon is burned, which suggests that oxygen isn't present.)
moles of C in the compound: 44 g x 1 mol C/mole and 27.8 g CO2 x 1 mol CO2. = 0.632 moles of carbon
9.96 g H2O x 1 mol H2O/18 g x 2 mol H/mol H2O Equals 1.11 moles of H in the molecule.
We can calculate the mass and verify that it adds up to 8.70 g to rule out the presence of oxygen in the original molecule.
mass C=0.632 mol C times 12 g/mol equals 7.58 g C
H mass equals 1.11 mol H = 1.11 g H x 1 g/mol
Total mass Equals 8.69 g.
We may divide both by the lowest value (0.632), which will give us the lowest whole number of moles.
0.632/0.632 = 1.0 moles C
H = 1.11/0.632 = 1.75 moles.
Now we can multiply them by 4 to get H's whole number, which is-
C = 4 moles.
H = 7 moles
∴Empirical equation: C4H7
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BUFFERS BUFFERS AND BUFFER CAPACI
INTRODUCTION
LABORATORY SIMULATION
How to calculate the pH of a buffer solution
1) Calculate the pH of a solution prepared by dissolving 2.05 g of sodium acetate, CH3COONa, in 50.0 mL of 0.25 Macetic
acid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. K₂ of CH3COOH is
1.75 X 10.5.
pH =
The pH of the solution prepared by dissolving sodium acetate and acetic acid is 5.05.
Given -
CH3COONa = 2.05g
So, Number of moles of CH3COONa = 2.05 / 82.03 = 0.024
∴ Molarity = 0.024 / 0.05
= 0.48 M
and CH3COOH = 0.25 M
pH of buffer - The equilibrium constant (Ka) of the weak acid and the ratio of weak base (A-) to weak acid (HA) in solution are the two variables that determine the pH of a buffer.
1) Various weak acids have different equilibrium constants (Ka).
2) Ka tells us how much of the HA in the solution will be dissolved into H+ and A-.
According to the question -
CH3COOH ⇄ CH3COO⁻ (aq) + H⁻ (aq)
Ka = [CH3COO⁻] [H⁺] / {CH3COOH]
[H⁺] = Ka x [CH3COOH] / [CH3COO⁻]
[H⁺] = 1.75 x 10⁻⁵ [0.25] / [ 0.48]
[H⁺] = 0.43 x 10⁻⁵ / 0.48
= 0.89 x 10⁻⁵
pH = -log [H⁺]
∴ pH = - log [0.89 x 10⁻⁵]
pH = 5.05
Hence, the pH of the solution will be 5.05.
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which substance can be broken down by a chemical change ?
1) magnesium
2) manganese
3) mercury
4) methanol
Answer:
Methanol
Explanation:
Methanol is a compound and compounds are pure substances formed by the combination of elements; they can be decomposed by ordinary chemical means. The other answers are incorrect because they are elements and elements consists of only one kind of atom and they cannot be broken down into a simpler type of matter by either physical or chemical means.
What are the products formed when an acid reacts with a metal carbonate?
The enthalpy of vaporization for methanol is 35.2 kJ/mol. Methanol has a vapor pressure of 101.3
kPa at 64.7 °C. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at
57.2 °C? Give your answer in kPa, to the first decimal point.
The vapour pressure of methanol at 57.2°C is 66.3 kpa.
The enthalpy of vaporization for methanol,
=35.2 kj/mol = 35.2×1000 j/mol
at temperature T2= 64.7°C = 64.7 + 273 = 337.4 K
temperature T1 = 57.2 °C = 57.2 + 273 = 330.2 K
vapour pressure P2 = 101.3 kpa
vapor pressure P1 = ?
Clausius - Clapeyron equation is geven as :
ln ( P1 / P2 ) = ΔH / R ( 1 / T2 - 1 / T1 )
ln ( P1 / 101.3 kpa ) = 35200 j/mol / 8.314 j/Kmol × ( 1/337.4 K - 1/330.2 K)
ln ( P1 / 101.3 kpa) = [tex]e^{-0.423}[/tex]
P1 / 101.3 kpa = 0.655
P1 = 0.655 × 101.3 kpa
P1 = 66.3 kpa
Hence, The vapor pressure at 57.2°C is 66.3 kpa.
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I forgot how to do fragment sentences
Answer:
A sentence fragment is a sentence that is missing either its subject or its main verb.
Explanation:
Incorrect: Went to the store yesterday.
Incorrect: After the classes, the library. My life nowadays
How many moles at gold (Au) atoms are in 87-89 of Au
The calculated answer is 1.99 x 10/21 x Au
Gold has a molar mass of 196.96657 g/m. The mass of 1 mole of gold atoms, which is commonly given in grams, is referred to as the molar mass of gold.
Divide the specified mass by the molar mass of gold, which is 196.966569 g/mol, to determine the moles.
The term "atomic weight" refers to an atom's overall weight. With the addition of a small amount from the electrons, it is roughly equal to the sum of the protons and neutrons. The number of neutrons an atom has has a significant impact on the stability of the nucleus and, consequently, the radioactivity of the atom. The average mass of an element's atoms expressed in atomic mass units is known as its atomic mass (amu, also known as daltons, D). Gold has a molar mass of 196.96657 g/m. The mass of 1 mole of gold atoms, which is commonly given in grams, is referred to as the molar mass of gold.
Divide the specified mass by the molar mass of gold, which is 196.966569 g/mol, to determine the moles.
(Atomic weight in g/mol on the periodic table).
0.00330 mol = 0.650 g Au = 196.966569 gmol Au Gold atoms 1 mole of atoms is equal to 6.02210/23 atoms.
double the estimated mol Au by several.
6.022×10/23 atoms1mol.
0.00330mol Au = 6.022 x 10/23 x 1 mol
= 1.99 x 10/21 x Au
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Because of increasing evidence of damage to the ozone layer, chlorofluorocarbon (CFC) production was banned in 1996. However, many older cars still have air conditioners that use CFC-12 (CF2Cl2). These air conditioners are recharged from stockpiled supplies of CFC-12. Suppose that 100 million automobiles each contain 1.0 kg of CFC-12 and leak 24 % of their CFC-12 into the atmosphere per year.
How much chlorine, in kg , is added to the atmosphere each year due to these air conditioners?
The mass of chlorine is 9.9 * 10^6 kg.
Total mass of CFC = M(CFC) * number of cars = 1.0 kg * 1.00 * 10^8
= 1.00 * 10^8 kg
Mass of the leaked CFC ;
% leaked = m(leaked) *100 / m(total)
m(leaked) = o.24 * 1.00 * 10^8 kg
m(leaked) = 2.4 * 10^6 kg
Now no. of moles of CFC:
n(CFC) = m(CFC) / MM(CFC) = 2.4 * 10^9 g / 170.92 g/mol = 0.014 * 10^9 mol
= 1.4 * 10^7 mol
No. of moles of chlorine ;
n(Cl) = n(CFC)*2
= 1.4 * 10^7 mol *2
= 2.8 * 10^7 mol
Now at last the mass of the chlorine :
m(Cl) = n(Cl) * MM(Cl)
= 2.8 * 10^7 mol * 35.453 g/mol
= 99.2 * 10^10 g
Mass of Cl = 9.9 * 10^6 Kg
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Cs(s)+HBr(aq)= what is the answer
Explanation:
Cs(s)+HBr(aq)=H(g)+CsBr(aq)
6) Which of the following statements does not describe a physical property of chlorine?
A) Chlorine combines with sodium to form table salt.
B) The color of chorine gas is green.
C) The density of chlorine gas at standard temperature and pressure is 3.17 g/L.
D) The freezing point of chlorine is -101°C.
Answer:
the answer is A. Chlorine combines with sodium to form table salt
Part A: Write a balanced nuclear equation for the fission reactions below.
1. Californium-252 undergoes spontaneous fission (no neutron collision) to produce xenon-140,
ruthenium-108, and an unknown number of neutrons. Write a balanced nuclear equation that
includes the correct number of neutrons produced. ( this is earth and space btw a form of science)
The spontaneous fission reaction is:
₉₈ Cf ²⁵² → ₅₄ Xe ¹⁵⁰+ ₄₄ Ru ¹⁰⁸ +4 ₀n¹
Given,
Californium-252 undergoes spontaneous fission to produce xenon-140, ruthenium-108, and an unknown number of neutrons.
The Nuclear reaction is
₉₈ Cf ²⁵² → ₅₄ Xe ¹⁵⁰+ ₄₄ Ru ¹⁰⁸ +4 ₀n¹
Spontaneous fission reaction always happens in heavier elements, that is the element with atomic numbers more than 89 and mass numbers greater than 230.
The nuclei of these elements are unstable and decay by radioactive decay in different forms like beta decay, positron, etc. At some percentage of the time, they decay by spontaneous fission.
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Name the following covalent compounds
The atoms of covalent compounds are connected by covalent bonds. Sharing one or more pairs of valence electrons results in the formation of a covalent bond.
How to name covalent compound-
1. Give the elemental name of the non-metal that is on the periodic table that is farthest to the left.
2. Give the other non-metal's name and a suffix ending, such as -ide.
3. To denote the quantity of an element in a molecule, use the prefixes mono-, di-, tri-, etc.
4. Mono is understood and not written if it is the initial prefix.
Names of covalent compounds according to the question-
1. CO₂ - Carbon Dioxide
2. CO - Carbon Monoxide
3. N₂O₄ - Dinitrogen tetraoxide
4. PCl₃ - Phosphorus trichloride
5. CBr₄ - Tetra bromomethane
6. P₂O₅ - Phosphorus pentoxide
7. PCl₅ - Phosphorus pentachloride
8. NF₃ - Nitrogen Trifluoride
9. SO₂ - Sulfur dioxide
10. SO₃ - Sulfur trioxide
11. SCl₆ - sulfur hexachloride
12. SiCl₄ - Silicon tetrachloride
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C. In what orbital is an electron located that has quantum numbers of: n = 5, 1= 0, m₁ = 0, ms=-1? (1
point)
The electron is in the 5s orbital.
What is an orbital?The term orbital is the region in space where there is a high probability of finding an electron. There are four quantum numbers that define an orbital and these are;
Principal quantum number
Orbital quantum number
Spin quantum number
Magnetic quantum number
Let us now look at the sets of quantum numbers as given, a look at the principal and the orbital quantum numbers (m and l) tell us that the electron is in the 5s orbital.
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what is an hydroxide solution commonly used in labs
sodium hydroxide (NAOH) a.k.a causic soda commonly used in labs.
MOLAR MASS of naoh is 39.997 g/mol.It is dry hard brittle, white stocks in large fused opaque white-stone like masses having a dense crystalline fracture in small pellets/flakes.it is very delinquent and one of the strongest base , acts as a saponifying agent which is used in industrial purposes.It is an important industrial chemical as it is used on its own and is also used as a raw material in the production of other chemicals. The basis for the production of sodium hydroxide is the electrolysis of brine.To know more about sodium hydroxide visit :
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Answer: Sodium hydroxide (NaOH), also known as caustic soda or lye, is a highly versatile substance used in a variety of manufacturing processes.
Explanation: For Example: Sodium hydroxide is also the most common base used in chemical laboratories, being able to test for quite a number of cations (this is called Qualitative Inorganic Analysis), as well as to provide alkaline mediums for some reactions that need it, such as the Biuret test.
the ph of a 0.04 M Calcium hydroxide is
The pH of a 0.04 M Calcium hydroxide solution is 1.39.
How to calculate pH?pH refers to the degree of alkalinity and acidity of a solution.
It is the quantitative measure of the acidity or basicity of a solution. The term "pH" is used in chemistry to denote the values of the concentration of the hydrogen ion {H+}.
The pH scale runs from 0 to 14 where; a value of seven is considered neutral, less than seven acidic, and greater than seven basic.
To calculate the pH of an aqueous solution, one needs to know the concentration of the hydronium ion in moles per liter (molarity).
The pH of a solution can be calculated as the negative logarithm of a concentration or molarity as follows:
pH = - log {H+}
According to this question, a 0.04 M calcium hydroxide is given. The pH can be calculated as follows:
pH = - log {0.04}
pH = 1.39
Therefore, the pH of a 0.04 M calcium hydroxide solution is 1.39.
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How would you prepare 10mL of a 0.25% m/V HCI solution if 1% m/V HCI was available? How much 1% m/V HCI is needed? How much distilled water is used?
We need to add 7.5 mL of distilled water to the 1% m/V HCI solution available.
What is the volume needed?The percent concentration is one of the common units of concentration that could be used to describe the amount of substance. Let us now try to obtain the volume needed.
Given that;
C1V1=C2V2
C1 = initial concentration
V1 = Volume of the stock solution
C2 = final concentration
V2 = volume of the new solution
Thus;
V1 = C2V2/C1
V1 = 0.25 * 10/1
= 2.5 mL
Hence, the volume of distilled water needed = 10mL - 2.5 mL
= 7.5 mL
Thus, we need to add 7.5 mL of distilled water to the 1% m/V HCI solution available.
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