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3 years ago

What is the ratio of blue-tailed to orange-talied critter pup?

Chemistry

1 answer:

exis [7]3 years ago

4 0

Blue to orange
blue:orange
$\frac{blue}{orange}$

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The basic building blocks of proteins are: O amino acids O aldehydes O amides ODNA

Sonja [21]

Amino acids are the basic parts of proteins

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3 years ago

Read 2 more answers

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

How many moles of water h2o are present in 75.0 g h2o?

nikklg [1K]

4.17 moles. Good luck! :)

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3 years ago

Substances that are less dense than water will_____ and substances with a density that is greater than 1.00 g/ml will____

Stells [14]

Substances that are less dense than water will FLOAT and substances with a density that is greater than 1.00g/ml will SINK.

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3 years ago

74. Which two temperatures are identical for the same substance?

AURORKA [14]

Answer:

(A) melting point, freezing point.

Explanation:

For the same substance: when matter is transitioning from solid to liquid (melting) or liquid to solid (freezing), its temperature is fixed at the melting/freezing point, which is the same temperature.

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3 years ago