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2 years ago

The reaction described in part a required 3.65 l of sodium chloride. what is the concentration of this sodium chloride solution?

1 answer:

7 0

Chemical reaction: AgNO₃ + NaCl → AgCl + NaNO₃.
Missing question: <span>Part A:What mass of silver chloride can be produced from 1.40 L of a 0.180 M solution of silver nitrate?
V(</span>AgNO₃) = 1,4 L.
c(AgNO₃) = 0,180 M = 0,180 mol/L.
V(NaCl) = 3,65 L.
n(AgNO₃) = V(AgNO₃) · c(AgNO₃).
n(AgNO₃) = 1,4 L · 0,180 mol/L.
n(AgNO₃) = 0,252 mol.
From chemical reaction: n(AgNO₃) : n(NaCl) = 1 : 1.
n(NaCl) = 0,252 mol.
c(NaCl) = 0,252 mol ÷ 3,65 L.
c(NaCl) = 0,069 mol/L

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3 years ago

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The compound is basic in nature but the concentration substance is corrosive.

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1 year ago

A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0

Maurinko [17]

<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

Boyle's Law: P₁V₁ = P₂V₂

P₁ is pressure 1
V₁ is volume 1
P₂ is pressure 2
V₂ is volume 2

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
[Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
[Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
[Pressure] Rewrite: P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

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2 years ago

30.5 g of sodium metal reacts with chlorine gas to produce sodium chloride. How much (in grams) chlorine gas must react with thi

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Answer:

Mass of chlorine = 47.22 g

Explanation:

Given data:

Mass of sodium = 30.5 g

Mass of chlorine= ?

Solution:

Chemical equation:

2Na + Cl₂ → 2NaCl

Number of moles of Na:

Number of moles = mass/molar mass

Number of moles = 30.5g/ 23 g/mol

Number of moles = 1.33 mol

Now we will compare the moles of Cl ₂ with Na from balance chemical equation.

Na : Cl ₂

2 : 1

1.33 : 1/2×1.33 = 0.665 mol

Mass of chlorine gas:

Mass = number of moles × molar mass

Mass = 0.665 mol × 71 g/mol

Mass = 47.22 g

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