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Andreas93 [3]
3 years ago
15

Name the salt produced if sodium carbonate reacts with dilute nitric acid ? Write the equation.

Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
5 0

Answer:

2HNO3 +Na2CO3 → CO2 + H2O + 2NaNO3

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Negatively charged particles in atom - Electrons
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Which of the following is the best hypothesis?
torisob [31]

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b

Explanation:

i think it is b

8 0
3 years ago
The ksp of pbi2 is 1.4 x 10-8. what is the molar solubility of lead(ii iodide in a solution of 0.400 m sodium iodide?
tino4ka555 [31]
The solubility product of a substance us calculated by the product of the concentration of the dissociated ions in the solution raise to the stoichiometric coefficient of the ions. Therefore, we need the dissociation reaction. For this, it will have the reaction:

PbI2 = Pb^2+ + 2I-

We solve as follows:

Ksp = [Pb2+][I-]^2 = <span>1.4 x 10-8
</span><span>1.4 x 10-8 = x(2x)^2
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x = 1.5x10^-3 M

The molar solubility would be </span>1.5x10^-3 M.
8 0
3 years ago
When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be
Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH  = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL  +  50.mL  = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH_{neutralization}

ΔH_{neutralization} = -q_soln / mole of water produced

so we substitute

ΔH_{neutralization} = -( 4076.68 J ) / 0.0500 mol  

ΔH_{neutralization} = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0
3 years ago
A saturated hydrocarbon is an organic compound which contains hydrogen and carbon joined _____.
Leya [2.2K]
B) with single bonds..... Hope it helps, Have a nice day :)
8 0
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Read 2 more answers
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