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fenix001 [56]
3 years ago
12

On a balance, you have beakers of AgNO3AgNO3 solution and NaClNaCl solution. When mixed, they will form AgCl(s)AgCl(s). What wil

l happen to the mass of the new mixture?
Chemistry
1 answer:
statuscvo [17]3 years ago
6 0

Explanation:

According to the law of conservation of mass, mass can neither be created nor it can be destroyed. But it can be simply transformed from one form to another.

Therefore, when AgNO_{3} is added to NaCl then the compound formed will have same mass as that of reactants.

       AgNO_{3} + NaCl \rightarrow AgCl + NaNO_{3}

Total mass of reactants is (169.87 + 58.44) g/mol = 228.31 g/mol

Total mass of products is (143.32 + 84.99) g/mol = 228.31 g/mol

Thus, we can conclude that mass of the new mixture will stay the same.

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Answer:

89.6L

Explanation:

1mole of any gas occupies 22.4L. This simply means that,

1mole of CO2 occupies 22.4L at stp.

Therefore, 4moles of CO2 will occupy = 4 x 22.4 = 89.6L

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4 0
4 years ago
What amount of heat is required to raise the temperature of 350 grams of copper to cause a 25°C change? The specific heat of cop
jarptica [38.1K]
The heat required to raise the temperature to a specific temperature change of a sample is related to the specific heat capacity of the substance.  In this case, the heat can be calculated through mCpΔT = 350 g * 0.39 J/g C *25 C. This is equal to 3412. 5 Joules. Closest answer is C.
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4 years ago
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In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
3 years ago
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