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goldfiish [28.3K]
3 years ago
9

You have 7.86 x 10^23 molecules of NaCl. This would be equal to ___ grams of NaCl.

Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0

7.86 \times 10 {}^{23} atoms \\  \frac{7.86}{6.022}  \times  \frac{10 {}^{23} }{ {10}^{23} }   = 1.305\: moles

1.305 × (23+35.5) = 76.34 grams

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on a hot summer day, water droplets often form on the outside of a cold glass. the water droplets form when water vapor in the a
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this is known as condensation when cold and warm temperatures meet


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A student mixes a 10.0 ml sample of 1.0 m naoh(aq) with a 10.0 ml sample of 1.0 m hcl(aq) in a polystyrene container. the temper
AleksAgata [21]
Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
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             = (10 + 10) (4.184) (26-20) = 502.08 J
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5 0
3 years ago
How many grams do 6.534e+24 molecules of phosphoric acid weigh?
fredd [130]

Answer:

1,063 grams H₃PO₄

Explanation:

To find the mass of phosphoric acid (H₃PO₄), you should (1) convert molecules to moles (via Avogadro's number) and then (2) convert moles to grams (via molar mass from periodic table).

Molar Mass (H₃PO₄): 3(1.008 g/mol) + 30.974 g/mol + 4(15.998 g/mol)

Molar Mas (H₃PO₄): 97.99 g/mol

6.534 x 10²⁴ molecules H₃PO₄                       1 mole                         97.99 g
---------------------------------------------  x  -------------------------------------  x  --------------
                                                            6.022 x 10²³ molecules          1 mole

= 1,063 grams H₃PO₄

3 0
2 years ago
The balanced redox reactions for the sequential reduction of vanadium are given below.
Minchanka [31]

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)

2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)

On adding above equations, we get the following combined equation.

2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn

We need 0.0392 grams of Zn metal to completely reduce vanadium.

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3 years ago
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