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3 years ago

What is Kb for CH3NH2(aq) + H2O(1) CH3NH3(aq) + OH (aq)?

Chemistry

1 answer:

statuscvo [17]3 years ago

8 0

Answer:

Kb for CH₃NH₂ (methylamine) is 4.4 × 10⁻⁴

Hope that helps.

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Calculate the number of grams of oxygen required to convert 48.0 g of glucose to co2 and h2o.

worty [1.4K]

I believe that the balanced chemical reaction is:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

So the number of grams of oxygen required is:

mass O2 required = 48 g C6H12O6 * (1 mole C6H12O6 / 180.16 g) * (6 mole O2 / 1 mole C6H12O6) * (32 grams O2 / 1 mole)

<span>mass O2 required = 51.15 grams</span>

4 0

3 years ago

I need some help understanding this.

otez555 [7]

Answer:

c^21, t^729, CH^56>748, do this to calculate the moceluar mass

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2 years ago

What is the molarity of a solution that contains 11.6 moles of LiNO3 in 13.7 liters of solution?

Tatiana [17]

The molarity of this should be around 0.846

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Formula: M= mols/liters

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So… M= 11.6/13.7⇒ 0.846

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6 0

1 year ago

Need help!!! Also don't joke around please

uysha [10]

47.867 u jdjsjsnsnnajjaisjsb

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2 years ago

Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to conv

poizon [28]

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

$Q_{1} = mC_{1} \Delta T_{1}$

Putting the given values into the above equation as follows.

$Q_{1} = mC_{1} \Delta T_{1}$

= $27.3 g \times 1.70 J/g K \times 41$

= 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

$Q_{2}$ = energy required = $mL_{v}$

$L_{v}$ = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

$Q_{2}$ = $mL_{v}$

= $27.3 \times 444$

= 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

$Q_{3} = mC_{2} \Delta T_{2}$

Value of $C_{2}$ = 1.06 J/g, $\Delta T_{2}$ = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

$Q_{3} = mC_{2} \Delta T_{2}$

= $27.3 g \times 1.06 J/g \times 34 K$

= 983.892 J

Therefore, net heat required will be calculated as follows.

Q = $Q_{1} + Q_{2} + Q_{3}$

= 1902.81 J + 12121.2 J + 983.892 J

= 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

5 0

2 years ago