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pochemuha
3 years ago
5

Can someone help me please.

Chemistry
1 answer:
Vika [28.1K]3 years ago
4 0

Answer:

ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

Explanation:

ccccccccccccccccccccccccccc

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what is the number of each type of atom on the right side of the equation 2Na3PO4(aq)+2CoCl2(aq)→2Co3(PO4)2(s)+6NaCl(aq)
Andre45 [30]

6 atoms of Cobalt.

4 atoms to Phosphate (PO4).

6 atoms to sodium.

6 atoms to Chlorine.

The coefficient is one factor that tells you how many atoms go to each atom/element, however, the subscript also influences this.

If the compound has parenthesis, then the subscript within the parenthesis remains untouched and does not affect the atoms.

Basically, to calculate the atoms is multiply the coefficient (number in front of the atom) by the subscript attached to the atom. If it's a compound such as Co3PO4, then make sure the coefficient affects the second atom as well.

Hope this helps!

3 0
3 years ago
Read 2 more answers
Calculate the mass, in grams, of cucl2 (mw = 134.452 g/mol) required to prepare 250.0 ml of a 6.11 % w/v cu2 (mw = 63.546 g/mol)
Sever21 [200]

6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution

therefore,  250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g

# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles

1 mole of CuCl2 contain 1 mole of Cu2+ ion

Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2

Molar mass of CuCl2 = 134.452 g/mole

The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams

6 0
3 years ago
Calculate the molarity of H3PO4 when you added 57.3 g into 3,820 mL of water
motikmotik

Answer:

0.153M

Explanation:

57.3/97.994 (molar mass)=0.585 moles of H3PO4

.0585/3.820L=0.153M

5 0
4 years ago
When acids react with water, ions are released which then combine with water molecules to form .
mestny [16]
When acids react with water, H ions are released which then combine with water molecules to form H₃O⁺
7 0
4 years ago
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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

5 0
3 years ago
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