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Doss [256]
3 years ago
9

Calculate the mass in 4.05*10^22 molecules of calcium phosphate

Chemistry
1 answer:
Marizza181 [45]3 years ago
8 0

Answer:

m = 20.9 g.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:

m=4.05x10^{22}molecules*\frac{1mol}{6.022x10^{23}}*\frac{310.18g}{1mol}\\\\m=20.9g

Regards!

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3 0
3 years ago
The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
3 years ago
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2 years ago
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8 0
3 years ago
8.310x10^2 – 7.210x10^1<br><br>[?]x10^[?]​
alisha [4.7K]

Answer:

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Explanation:

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4) (8.310\times 10^{1}-7.210)\times 10^{1} Distributive property

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6) 75.89\times 10^{1} Subtraction.

7) (7.589\times 10^{1})\times 10^{1} Multiplication/Associative property

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