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3 years ago

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Cooling the concentrated ammonia and hydrochloric acid before carrying out the experiment increased the time taken for the white

cloud to form

1 answer:

Crazy boy [7]3 years ago

8 0

Hope it helps you :)

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Am I incorrect ?<br> Or nah?

Mademuasel [1]

I think so... I'm currently learning this too but you should be correct

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3 years ago

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For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th

Stolb23 [73]

Given:

Diprotic weak acid H2A:

Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m

Balanced chemical equation:

H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

solve for x and determine the concentration at equilibrium.

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3 years ago

Why is it difficult for countries to reduce emissions of carbon dioxide

Natalka [10]

Answer:

Curbing global carbon dioxide emissions has been a challenge, primarily because they are being driven higher by countries with low per capita emissions.

Explanation:

"just trust me bro"

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3 years ago

In the compound H2O, the electrons in the bonds are unequally shared between oxygen and hydrogen, forming ____. (1 point)

Aloiza [94]

<span>polar bonds due to high difference . In electronegativity of oxygen and hydrogen.</span>

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3 years ago

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The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

3 years ago