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3 years ago

The standard unit for measuring volume is the _____.

Chemistry

1 answer:

Otrada [13]3 years ago

4 0

The answer is B I hope this helps you

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Explain how to use the periodic table to deduce the number of protons, neutrons and electrons of an atom of a specific element

Ksju [112]

Answer:

See explanation

Explanation:

The periodic table shows the atomic number and mass number of each element.

We know that the atomic number shows;

The number of protons in the nucleus of the atom
The number of electrons in the neutral atom of the element.

So we obtain the number of protons and electrons by looking at the atomic number shown in the periodic table.

We also know that;

Mass number = Number of protons + number of neutrons

Since number of protons = atomic number of the atom

Number of neutrons = Mass number - atomic number

Hence we obtain the number of protons by subtracting the atomic number from the mass number given in the periodic table.

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4 years ago

What would be the most appropriate metric unit to describe the mass of a person? A)grams

ValentinkaMS [17]

The most appropriate and most commonly used metric when describing a person's mass is D.) kilograms

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4 years ago

Read 2 more answers

At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.

natka813 [3]

Explanation:

The given data is as follows.

T = 298 K, $\Delta H^{o}$ = -5645 kJ/mol

$\Delta G^{o}$ = -5798 kJ/mol

Relation between $\Delta H$ and $\Delta G$ are as follows.

$\Delta G^{o}$ = $\Delta H^{o} - T \Delta S^{o}$

-5798 kJ/mol = -5645 kJ/mol - $298 \times \Delta S^{o}$

-153 kJ/mol = - $298 \times \Delta S^{o}$

$\Delta S^{o}$ = 0.513 kJ/mol K

Now, temperature is $37^{o}C$ = (37 + 273) K = 310 K

Since, $\Delta G$ = $\Delta H^{o} - T \Delta S^{o}$

= $-5645 kJ/mol - 310 K \times 0.513 kJ/mol K$

= (-5645 kJ/mol - 159.03 kJ/mol)

= -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

$\Delta G = \Delta G_{310 K} - \Delta G_{298 K}$

= -5804.03 kJ/mol - (-5798 kJ/mol)

= -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

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3 years ago