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3 years ago

A mineral has a mass of

Chemistry

2 answers:

ZanzabumX [31]3 years ago

7 0

192 g because you would have to multiply 24 by 8

Mila [183]3 years ago

4 0

Answer:

Explanation:

density= mass/ volume

=24/8

=3kg/meter cube

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Explain the units and how to solve the following metric conversion then solve the metric conversion by filling in the blank 12KM

LiRa [457]

Answer:- 12 km = 12000 m

Solution:- It's a metric unit conversion where we are asked to convert 12 km to m where km stands for kilometer and m stands for meter.

In metric conversions, kilo means 1000.

So, 1 km = 1000 m

It means, we multiply the given km by 1000 to get the answer in m as:

$12km(\frac{1000m}{1km})$

= 12000 m

Hence, 12 km = 12000 m.

5 0

3 years ago

What is the concentration of H+ ions at a pH = 11?

natta225 [31]

Answer:

$\huge 1 × {10}^{-11} \: \: M$

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ {H}^{+} ]$

Since we are finding the H+ ions we find the antilog of the pH

So we have

$11 = - log({H}^{+}) \\ {H}^{+} = {10}^{ - 11}$

We have the final answer as

$1 × {10}^{-11} \: \: M$

Hope this helps you

6 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

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

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

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3 years ago

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Hellpppppp! Please!!!!

lana [24]

the answer is :the concentration the chemicals in the solution because the more ions there are in the solution the higher the conductivity also the more ions there are in the solution the stronger the electrolyte

6 0

3 years ago

Which of the following statements is true? A) The iron atom is best considered to be a solid. B) The mercury atom is best consid

shepuryov [24]

I believe the correct response would be D. At least 2 of the above statements are correct.

3 0

3 years ago

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