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3 years ago

Convert the following word equations into chemical equations using the correct

Chemistry

1 answer:

Alex17521 [72]3 years ago

7 0

Answer:

Below.

Explanation:

1. Al2(SO4)^3 (aq) + 6LiOH aql) ----> 3Li2SO4 (aq) + 2Al(OH)^3 (s)

2. (NH4)2CO3 (aq) + MgCl2 (aq) ----> 2(NH4)Cl (aq) + MgCO3 (s).

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How many moles of Boron (B) are in 5.03 x 1024 B atoms?

goldfiish [28.3K]

Hey there!:

Number of moles = ( number of atoms / 6.023*10²³ atoms )

given number of atoms = 5.03*10²⁴

Therefore:

Number of moles B = 5.03*10²⁴ / 6.023*10²³

Number of moles B = 8.35 moles

Hope that helps!

3 0

2 years ago

Question 2 Unsaved Which of the following chemical formulas shows a covalent compound? a)CO2 b)MgO c)NaCl d)HBr

zzz [600]

It would be D
Because a covelant compound forms when 2 non metal atoms bond

6 0

3 years ago

Read 2 more answers

The less energy a wave has, the greater its amplitude. True or False

S_A_V [24]

Answer:

true

Explanation:

the more energy a wave has the greater the amplitude is

6 0

2 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$