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Alecsey [184]
2 years ago
7

Determine the redox reaction represented by the following cell notation: Mg(s) | Mg2 (aq) || Cu2 (aq) | Cu(s)

Chemistry
1 answer:
mrs_skeptik [129]2 years ago
6 0
Oxidation: Mg (s) —> Mg2+ (aq) + 2e
reduction: Cu2+ (aq) +2e —> Cu(s)

redox reaction: Mg(s) + Cu2+(aq)—> Mg2+(aq) + Cu(s)
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The weak interactions between atoms that keep atoms near each other but don't tightly bind them together are called
makkiz [27]

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Cohesive and adhesive forces.

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7 0
3 years ago
Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?
WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V  

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

6 0
2 years ago
____H2 + ____O2 ---&gt; ____H20<br><br> 2, 1, 1<br> 2, 2, 1<br> 2, 1, 2<br> 1, 2, 2
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6 0
3 years ago
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Answer and work for this problem
MArishka [77]
We can write the balanced equation for the synthesis reaction as 
     H2(g) + Cl2(g) → 2HCl(g)

We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
     mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) * 
                           (2.02 g H2 / 1 mol H2)                        
                        = 4.056 g H2

We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
     mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
                             (70.91 g Cl2 / 1 mol Cl2)
                          = 142.4 g Cl2 

Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
6 0
3 years ago
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