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pochemuha
3 years ago
14

Given the following reaction:

Chemistry
1 answer:
Semenov [28]3 years ago
7 0

Answer:

The answer to your question is 0.35 moles of ZnCl₂

Explanation:

Data

moles of ZnCl₂ = ?

mass of Zn = 23 g

mass of CuCl₂ = excess

Balanced chemical reaction

             Zn + CuCl₂   ⇒   ZnCl₂  +  Cu

Process

1.- Convert the mass of Zn to moles

-Look for the atomic mass of Zn

Atomic mass = 65.4 g

-Convert the grams to moles

            65.4 g -------------- 1 mol

             23 g  ----------------  x

               x = (23 x 1)/65.4

               x = 23 / 65.4

               x = 0.35 moles of Zn

2.- Calculate the moles of ZnCl₂ using proportions and cross multiplication, and the coefficients of the balanced chemical reaction.

              1 mol of Zn ---------------- 1 mol of ZnCl₂

             0.35 moles of Zn --------  x

              x = (0.35 x 1) / 1

              x = 0.35 moles of ZnCl₂

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2 years ago
If an ideal gas has a pressure of 2.97 atm, a temperature of 449 K, and has a volume of 58.35 L, how many miles of gas are in th
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N = ?

T = 449 K

V = 58.35 L

P =2.97

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

2.97 x 58.35 = n x 0.082 x 449

173.2995 = n x 36.818

n = 173.2995 / 36.818

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hope this helps!



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If 50 g of lead (of specific heat 0.11 kcal/kg ∙ C°) at 100°C is put into 75 g of water (of specific heat 1.0 kcal/kg ∙ C°) at 0
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Answer: The final temperature is 279.8K

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of lead = 50 g

m_2 = mass of water = 75 g

T_{final} = final temperature = ?

T_1 = temperature of lead = 100^oC=373K

T_2 = temperature of water = 0^oC=273K

c_1 = specific heat of lead = 0.11kcal/kg^0C

c_2 = specific heat of water= 1.0kcal/kg^0C

Now put all the given values in equation (1), we get

50\times 0.11\times (T_{final}-373)=-[75\times 1.0\times (T_{final}-273)]

T_{final}=279.8K

Therefore, the final temperature of the mixture will be 279.8 K.

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