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3 years ago

A as dhitssfghj plz help help

Chemistry

1 answer:

Alex Ar [27]3 years ago

6 0

I think B, don’t come at me if I’m wrong

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2 years ago

Calculate the volume of a 0.5M solution containing 20g of NaOH

Tju [1.3M]

Answer:

Explanation:

First, let us calculate the number of mole present in 20g of NaOH. This is illustrated below:

Mass = 20g

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Number of mole =?

Number of mole = Mass /Molar Mass

Number of mole of NaOH = 20/40 = 0.5mol

From the question given, we obtained the following data:

Molarity = 0.5M

Mole = 0.5mole

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.5/0.5

Volume = 1L

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3 years ago

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MrRissso [65]

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3 years ago

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