Which substance is an example of an element?

A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of t

ioda

Answer:

0.053 L is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

4 0

3 years ago

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You make a solution by putting 45.6g of iron lll carbonate into 167ml of water. What is it's molarity?

ludmilkaskok [199]

1. The molar mass of Fe2(CO3)3 is 291.72 g/mol. This means that 45.6 g is equivalent to 0.156 mol. Dividing by the 0.167 L of water gives a solution of 0.936 M.
2. Multiplying (0.672 M)(0.025 L) = 0.0168 mol. The molar mass of Ni(OH)2 is 92.71 g/mol, so multiplying by 0.0168 mol = 1.56 grams. Therefore you would need to dissolved 1.56 g of Ni(OH)2 into 25 mL of water.
3. Fe2(CO3)3 + Ni(OH)2 --> Fe(OH)3 + NiCO3Balancing: Fe2(CO3)3 + 3Ni(OH)2 --> 2Fe(OH)3 + 3NiCO3The reaction quotient is:[Fe(OH)3]^2 * [NiCO3]^3 / [Fe2(CO3)3][Ni(OH)2]^3= (0.05)^2 * (1.45)^3 / (0.936)(0.672)^3= 0.0268Since this is < 1, it implies that the reactants are favored at equilibrium.

4 0

3 years ago

What kind of reaction occurs when you mix aqueous solutions of barium sulfide and sulfuric acid? what kind of reaction occurs wh

My name is Ann [436]

The kind of reaction that occurs when you mix aqueous solutions of barium sulfide and sulfuric acid is a precipitation reaction.

<h3>Further Explanation</h3>

The chemical reaction between Ba(OH)2(aq) and H2SO4(aq) is given by;

Ba(OH)₂(aq) + H₂SO4(aq) --> BaSO₄(aq) + 2H₂O(l)

This is a type of precipitation reaction, where a precipitate is formed after the reaction, that is Barium sulfate.

<h3>Other types of reaction</h3><h3>Neutralization reactions </h3>

These are reactions that involve reacting acids and bases or alkali to form salt and water as the only products.
For example a reaction between sodium hydroxide and sulfuric acid.

NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l)

<h3>Displacement reactions</h3>

These are reactions in which a more reactive atom or ion displaces a less reactive ion from its salt.

Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)

<h3>Redox reactions </h3>

These are reactions that involve both reduction and oxidation occuring simultaneously durin a chemical reaction.
For example,

Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)

Magnesium atom undergoes oxidation while copper ions undergoes reduction.

<h3>Decomposition reactions</h3>

These are type of reactions that involves breakdown of a compound into its constituents elements.

For example decomposition of lead nitrate.

Pb(NO3)2(S) → PbO(s) + O2(g) + NO2(g)

Keywords: Precipitation

<h3>Learn more about: </h3>

Precipitation reaction: brainly.com/question/11194650
Examples of precipitation reactions: brainly.com/question/11194650
Neutralization reactions brainly.com/question/3243813

Level: High school

Subject: Chemistry

Topic: Chemical reactions

Sub-topic: Precipitation reactions

3 0

3 years ago

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Does the following compound contain a polyatomic ion? Na2SO4<br> Yes<br> No

seropon [69]

The answer is yes I believe so.

8 0

2 years ago

). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50

Nikitich [7]

Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C = 2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568 g C

Mass of H = 0.6955 g H₂O × (2.016 H/18.02 g H₂O) = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

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2. Convert these masses to moles.

Moles C = 0.5568 × 1/12.01 = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654 × 1/16.00 = 0.054 09

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3. Find the molar ratios.

Moles C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36 = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

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4. Multiply the ratios by a number to make them close to integers

C = 1 × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6 = 7.001

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5. Round the ratios to integers

C:H:O =6:10:7

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6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

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7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

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8. Divide the molecular mass by the empirical formula mass.

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

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9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

7 0

3 years ago