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3 years ago

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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3 years ago

What is the molarity of 20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of solution

KATRIN_1 [288]

Answer:

0.8M

Explanation:

CM=n/V

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3 years ago

How does an indicator differentiate between an acid and base?

Svet_ta [14]

Answer:

that is all I know sorry hope that may help you :)

Explanation:

Indicator is a substance which shows different colors in acidic and basic medium. Indicators can be natural (derived from natural sources) or artificial ( man-made) for example :

*Litmus is an indicator. Acid turns blue litmus into red while base turns red litmus into blue

* Turmeric solution does not show change in color (remains yellow) in acidic solution and turns red in basic solution.

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2 years ago

grapevine trimming and other aromatics can be wrapped in aluminum foil with a few holes poked in it to?

Juli2301 [7.4K]

I need a little more context but I believe you are correct

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3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago