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Roman55 [17]
2 years ago
13

Lead 2 nitrate reacts with sodium sulfate to form lead 2 sulfate and sodium nitrate. Balance the equation.

Chemistry
1 answer:
Mama L [17]2 years ago
3 0

Answer:

Explanation:

Pb(NO₃)+Na(SO₄)⇒Pb(SO₄)+Na(NO₃)

make them electrically balanced

Pb(NO₃)₂+Na₂(SO₄)⇒Pb(SO₄)+Na(NO₃)

balance

Pb(NO₃)₂+Na₂(SO₄)⇒Pb(SO₄)+2Na(NO₃)

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Who is generally credited with developing the process of using heat?
galina1969 [7]

Answer:

Robert Boyle

Explanation:

I think it's that

6 0
2 years ago
Calculate the Molarity when a 6.11 mL solution of 0.1 H2SO4 is diluted with 105.12 mL of water
barxatty [35]

Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same  

The dilution formula

 M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

M₁=0.1 M

V₁=6.11

V₂=105.12

\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M

5 0
3 years ago
Use the molar solubility 3.27×10−11m in pure water to calculate ksp for nis.
Ne4ueva [31]

Answer:

Ksp = 1.07x10⁻²¹

Explanation:

Molar solubility is defined as moles of solute can be dissolved in 1L.

Ksp for NiS is defined as:

NiS(s) ⇄ Ni²⁺(aq) + S²⁻(aq)

Ksp = [Ni²⁺] [S²⁻]

As molar solubility is 3.27x10⁻¹¹M, concentration of [Ni²⁺] and [S²⁻] is 3.27x10⁻¹¹M for both.

Replacing:

Ksp = [3.27x10⁻¹¹M] [3.27x10⁻¹¹M]

<em>Ksp = 1.07x10⁻²¹</em>

<em></em>

4 0
2 years ago
Gastric juice in the stomach contains pepsin and sulfuric acid.<br><br><br> TrueFalse
jonny [76]
False, pepsin and hydrochloric acid
5 0
2 years ago
What volume of water would be added to 16.5 ml of a 0.0813 m solution of sodium borate in order to get a 0.0200 m s?
raketka [301]
When diluting solutions from concentrated solutions the following formula can be used 
c1v1 = c2v2 
where c1 is concentration and v1 is volume of the concentrated solution 
and c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting these values 
0.0813 M x 16.5 mL = 0.0200 M x V
V = 67.1 mL
the volume of the diluted solution prepared is 67.1 mL.
the volume of water that should be added to get a final volume of 67.1 mL is (67.1 - 16.5 ) = 50.6 mL
a volume fo 50.6 mL should be added 
7 0
3 years ago
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