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2 years ago

Can yall answer the question on the image PLEASE

Chemistry

1 answer:

Brut [27]2 years ago

6 0

Answer is 67.8%

To find the percent yield the equation is actual yield over theoretical yield, and multiply by 100. In this problem it is 4.06996 / 6 = 0.678 x 100 = 67.8%

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What coefficient of h+ balances the atoms in this half-reaction h+ + mno2 → mn2+ + h2o?

kvv77 [185]

The answer is 4 im pretty sure

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3 years ago

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

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

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

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3 years ago

Which of these is true for natural selection but not artificial selection?

ElenaW [278]

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3 years ago

Hydrogen 3 has a half life of 12.32 years a sample of h-3 weighing 3.02 grams is left for 15.0 years what will the final weight

yawa3891 [41]

Answer:

The final mass of sample is 1.3 g.

Explanation:

Given data:

Half life of H-3 = 12.32 years

Amount left for 15.0 years = 3.02 g

Final amount = ?

Solution:

First all we will calculate the decay constant.

t₁/₂ = ln² /k

t₁/₂ =12.32 years

12.32 y = ln² /k

k = ln²/12.32 y

k = 0.05626 y⁻¹

Now we will find the original amount:

ln (A°/A) = Kt

ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y

ln (3.02 g/ A) = 0.8439

3.02 g/ A = e⁰°⁸⁴³⁹

3.02 g/ A = 2.33

A = 3.02 g/ 2.33

A = 1.3 g

The final mass of sample is 1.3 g.

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3 years ago

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How do scientist learn about the mantle and core of the earth

exis [7]

They drill too deep and find lava

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