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3 years ago

PLZ Help asap!!!!

Chemistry

1 answer:

Naily [24]3 years ago

5 0

Answer:

C. 9.70

Explanation:

Hello,

This problem is a simple dilution problem which utilizes the formula M1V1= M2V2 (where M is molarity, V is volume). I have attached a helpful image as well.

M1 V1 = M2 V2

(18) (x) = (0.35) (500)

x = 9.72mL which is close enough to option C.

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Question 5(Multiple Choice Worth 3 points)

Katarina [22]

Answer:

$\sqrt{ | {o2}^{2} | } \times \frac{?}{?} \sqrt[?]{?} \times \frac{?}{?} \sqrt[ |?| ]{?} \sqrt{?} \times \frac{?}{?}$

Explanation:

exactly

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3 years ago

What is an atomic number? An atomic mass?

mrs_skeptik [129]

Answer:

The atomic number is the number of the elements inside the periodic table and the mass is the weight or a number under the elements.

Explanation:

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3 years ago

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

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

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

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

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

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

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

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 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

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3 years ago

What is the ph of a solution labeled .30 trimethylamine k for trimethylamine is 7.42 x 10^4

valentina_108 [34]

Answer:the pH is 12

Explanation:

First We need to understand the structure of trimethylamine

Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline

C3H9N+ H2O --> C3H9NH + OH-

$k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}$

Then:

The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine

[OH-]= $\sqrt{ 0.3*7.42x10^{-4}}$

[OH-]=0.01

pH=14-(-log[OH-])

pH=12

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3 years ago

A sample of carbon-12 has a mass of 6.00 g. How many atoms of carbon-12 are in the sample?

Bingel [31]

Carbon molar mass=12
carbon mass in the above question =6
mole=number of atoms/6.02x10²³
6/12= number of atoms/6.02x10²³
number of atoms=1/2 x6.02x10²³
number of atoms= 6.02 x10²³/2
number of atoms=3.01x10²³

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3 years ago

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