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denis23 [38]
2 years ago
11

Plz help!!! This is timed!!!

Chemistry
1 answer:
pogonyaev2 years ago
7 0
58.7 %

Please correct me if I’m wrong. :)
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cuantos gramos de oxígeno necesarios para que reaccionen completamente 28 gramos de etanol (c2h5oh) en la reaccion de combustion
tigry1 [53]

Answer:

I dont speak this language sorry :(

Explanation:

4 0
3 years ago
( BRAINLIEST) Which of these statements best describes an Alkali? *
Bad White [126]

Answer:

soluble base

Explanation:

as alkali is a salt used as base

7 0
3 years ago
Read 2 more answers
A first-order reaction (A → B) has a half-life of 25 minutes. If the initial concentration of A is 0.900 M, what is the concentr
Oksanka [162]

Answer:

0.6749 M is the concentration of B after 50 minutes.

Explanation:

A → B

Half life of the reaction = t_{1/2}=25 minutes

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{25 min}=0.02772 min^{-1}

Initial concentration of A = [A]_o=0.900 M

Final concentration of A after 50 minutes = [A]=?

t = 50 minute

[A]=[A]_o\times e^{-kt}

[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

0.6749 M is the concentration of B after 50 minutes.

7 0
3 years ago
Fermeroben heres the messages COUGH
lord [1]

I completely understand your gripe. I also believe that you are the better option. It will only take time until she is yours, despite your parent's views. Keep it up

6 0
3 years ago
A certain liquid X has a normal boiling point of 133.60°C and a boiling point elevation constant Kb= 2.46°C kg mol^-1.Calculate
Afina-wow [57]

Answer:

136.63 °C

Explanation:

ΔTb=Tb solution - Tb pure

Where; Tb pure = 133.60°C

molar mass of solute = 121.14 g/mol

number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles

molality = 0.431 moles/350 * 10^-3 = 1.23 molal

Then;

ΔTb = Kb * m * i

Kb = 2.46°C kg mol^-1

m = 1.23 molal

i = 1

ΔTb = 2.46 * 1.23 * 1

ΔTb = 3.03 °C

Hence;

Tb solution = ΔTb + Tb pure

Tb solution = 3.03 °C + 133.60°C

Tb solution = 136.63 °C

6 0
3 years ago
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