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3 years ago

When we orbit the EARTH!DO WE HAVE ALL THE TIME TO OURSELVES...MHHHH

Chemistry

1 answer:

Sveta_85 [38]3 years ago

3 0

Answer

oh my.. ima put that down in my conspiracy theory book

Explanation:

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How do kidneys work with other body systems to maintain homeostasis?

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Kidneys work with other body systems to maintain homeostasis by working with other body systems such as the respiratory system.

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3 years ago

4. Solve the following heat flow problem, being sure to show all your work (you may either type your

Viktor [21]

Answer:

0.70 J/g.°C

Explanation:

Step 1: Given data

Mass of graphite (m): 402 g

Heat absorbed (Q): 1136 J

Initial temperature: 26°C

Final temperature: 30 °C

Specific heat of graphite (c): ?

Step 2: Calculate the specific heat of graphite

We will use the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 1136 J / 402 g × (30°C - 26°C)

c = 0.70 J/g.°C

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3 years ago

Which of the following is an oxidation? Select all that apply

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Answer:

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2 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

Please help me with these two questions

ANEK [815]

What is the question?

8 0

3 years ago

When we orbit the EARTH!DO WE HAVE ALL THE TIME TO OURSELVES...MHHHH​

When we orbit the EARTH!DO WE HAVE ALL THE TIME TO OURSELVES...MHHHH