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Angelina_Jolie [31]
3 years ago
9

Calculate the number of ammonia molecules in 3.9 g.

Chemistry
1 answer:
disa [49]3 years ago
6 0
•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol

1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:

(3.9 g)/ (17.03g/mol) = 0.22900763mols

Then convert the moles to molecules by multiplying it with Avogadro’s number:

Avogadro’s number: 6.022 x 10^23


0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
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The following reaction shows sodium carbonate reacting with calcium hydroxide.
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Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams


5 0
3 years ago
How many grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl?
neonofarm [45]
1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:

n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol

2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation

[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg

3) Transform mol to mass using the atomic mass:

10.4 mol Mg * 24.3 g/mol = 252.7 g of Mg.

4 0
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Which of the following is formed by intense heat and pressure over millions of years?
kumpel [21]

Answer:

B:Metamorphic

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8 0
3 years ago
Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re
bearhunter [10]

Answer:  28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}   

\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles

The balanced chemical equuation is:

4Al+3O_2\rightarrow 2Al_2O_3  

According to stoichiometry :

4 moles of Al produce == 2 moles of Al_2O_3

Thus 0.556 moles of Al will produce=\frac{2}{4}\times 0.556=0.278moles  of Al_2O_3

Mass of Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0
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