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2 years ago

Please someone help mee

Chemistry

1 answer:

Nikitich [7]2 years ago

8 0

Answer:

13.9kj/mol

Explanation:

$h = 891 -(393.5 +2*241.8)=1768.1=13.9kj/mol$

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Do equal volumes of the same gas at the same temperature and pressure contain equal numbers of molecules (number of moles)?

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Answer:Therefore, there must be a direct relationship between these volumes of gases and the number of molecules they contain. Avogadro's law says that: Equal volumes of different gaseous substances, measured under the same pressure and temperature conditions, contain the same number of molecules.

Explanation:I hope it works for you.

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3 years ago

Magma that is rich in feldspar and silica is called?

weeeeeb [17]

Answer:

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2 years ago

What is the atom inventory for the following equation after it is properly balanced? ____NaOH + ____CuCl2 Imported Asset ____NaC

faltersainse [42]

Answer:

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2;

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

Explanation:

NaOH + CuCL2 —> NaCl + Cu(OH)2

The balanced equation can be achieved by doing the following:

There are 2 oxygen and 2 hydrogen atom on the right side. This is balanced by putting 2 in front of NaOH as shown below:

2NaOH + CuCL2 —> NaCl + Cu(OH)2

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Now the equation is balanced.

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Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

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3 years ago

7.) Regarding molecular formula and empirical formula for a particular compound... (A) the subscripts in the empirical formula m

S_A_V [24]

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2 years ago

Read 2 more answers

a 10.99g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (define proportions), how many g

leonid [27]

Given :

A 10.99 g sample of NaBr contains 22.34% Na by mass.

To Find :

How many grams of sodium does a 9.77g sample of sodium bromine contain.

Solution :

By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.

Therefore , percentage of Na by mass in NaBr will be same for every amount .

Percentage of Na in 9.77 g NaBr is 22.34 % too .

Gram of Na = $9.77\times \dfrac{22.34}{100}=2.18\ g$ .

Hence , this is the required solution .

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2 years ago