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Norma-Jean [14]
2 years ago
8

A hydrate of zinc nitrate has the formula Zn(NO3)2 . xH2O. If the mass of 1 mol of anhydrous zinc nitrate is 63.67% of the mass

of 1 mol of the hydrate, what is the value of x?
Chemistry
1 answer:
weqwewe [10]2 years ago
6 0

MM Zn(NO₃)₂ = 189.36 g/mol

mass 1 mol Zn(NO₃)₂ = 189.36 g

mass hydrate = 100 / 63.67 x 189.36 = 297.409 g

mass 1 mol hydrate = 297.409 g

MM hydrate = 297.409 g/mol

MM hydrate = MM Zn(NO₃)₂ + MM xH₂O

297.409 = 189.36 + x(18)

x = 6

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a resource that cannot be replenished in a short period of time

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3 years ago
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Which of the following is true about the elements on the Periodic Table? There may be 1 or more correct answers.
nlexa [21]

Answer: 1. Block

2. True ( I'm unsure. He arranged it according to mass, but he is credited for the periodic table)

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Explanation:

4 0
2 years ago
Equation is balanced already
mojhsa [17]

Answer:

1.) 13 g C₄H₁₀

2.) 41 g CO₂

Explanation:

To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O

 48 g O₂             1 mole             2 moles C₄H₁₀            58.124 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  13 g C₄H₁₀

 48 g O₂             1 mole               8 moles CO₂            44.007 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  41 g CO₂

6 0
1 year ago
A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C
Mamont248 [21]

Answer:

T_{eq}=23.85^oC

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

\Delta H_{Cu}=-\Delta H_{H_2O}

Therefore the equilibrium temperature shows up as:

m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC

Best regards.

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