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3 years ago

Can you pls tell me the word equations for all these equations

Chemistry

1 answer:

Leokris [45]3 years ago

5 0

Answer:

Below

Explanation:

Balanced form;

$2C_6H_6 +15O_2 -> 12CO_2+6H_2O\\\\ Ca_3(PO_4)_2 +8C -> Ca_3P_2 +8CO\\\\2HNO_2+O_2 -> 2HNO_3\\\\Ca(OH)_2 + CO_2 -> CaC0_3+H_2O\\\\2K +Br_2 ->2KBr\\\\2NaOH+FeSO_4 -. Na_2SO_4 +Fe(OH)_2$

1.Benzene + Dioxygen = Carbon Dioxide + Water

2.Tricalcium phosphate +Carbon = Calcium phosphide + carbon monoxide

3.Nitrous acid react with oxygen to produce nitric acid.

4.This means that the carbon dioxide and limewater react to produce calcium carbonate and water.

5.Potassium react with bromine to produce potassium bromide

6. An aqueous solution of ferrous sulphate reacts with aqueous solution of sodium hydroxide to form a precipitate of ferrous hydroxide and sodium sulphate remains in the solution.

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1 year ago

.) What happens when Carbon dioxide gas is Collected down ward of water ?

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2 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

2 years ago

(8.6 1029) 7.4 X1029)

andrew11 [14]

Answer:

65563.914234

Explanation:

8.61029 x 7.4 x 1029

63.716146 x 1029

multiply

= 65563.914234

3 0

2 years ago

Read 2 more answers

Which of the following is true for an excess reactant?

kupik [55]

Answer:

C) It is the reactant that is left over after the reaction stops.

Explanation:

The excess reactant is the reactant that is left over after the reaction stops. The extent of the reaction is not determined by this reactant.

A limiting reactant is a reactant that is in short supply within a given reaction.

Such reactants determines the extent of chemical reaction.

Limiting reactants are used up in a chemical reaction.
The excess reactants remains unchanged after the reaction.

8 0

2 years ago

Can you pls tell me the word equations for all these equations​

Can you pls tell me the word equations for all these equations