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4 years ago

Click on the molecules that represent transpiration.

1 answer:

8 0

Transpiration is the progression of water inside a plant! So, the molecule representing transpiration is going to be good ol' H2O! =)

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3. How do we get mass from moles?

maria [59]

I think the answer is B

7 0

3 years ago

A buffer is prepared by dissolving 0.80 moles of NH3 and 0.80 moles of NH4Cl in 1.00 L of aqueous solution. If 0.10 mol of NaOH

Vikki [24]

Answer:

a. 10.54

Explanation:

reaction is

NH₃ + NaOH -----------------NH₄Cl + H₂O

0.10 mol NaOH will consume 0.10 mol NH₃ thereby decreasing the initial amount of moles NH₃ and increasing that of NH₄Cl

mol NH₃ = 0.80 - 010 = .70

mol NH₄Cl = 0.80 + .10 = 0.90

pH = pKₐ + log (( NH₃/NH₄Cl))

pH = 9.26 + log (( 0.90 + 0.70)) = 9.26 + 0.11 = 10.54

7 0

3 years ago

Calculate the volume of a sample of iron that has a density of 6.8 g/mL and a mass of 6780 mg.

Veronika [31]

Answer:

m= 6780mg = 6.78g = 6.8g

d= 6.8g/ml

V= m/d

V= 1ml

8 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

Anna [14]

Answer: The number of moles of ethanol after equilibrium is reached the second time is 11. moles.

Explanation:

We are given:

Initial moles of ethene = 34 moles

Initial moles of water vapor = 15 moles

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 34 15

At eqllm: 34-x 15-x x

We are given:

Equilibrium moles of ethene = 24 moles

Equilibrium moles of water vapor = 5 moles

Calculating for 'x'. we get:

$34-x=24\\\\x=10$

Volume of container = 100.0 L

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}$ .......(1)

$[CH_3CH_2OH]=\frac{10}{100}=0.1M$

$[CH_2=CH_2]=\frac{24}{100}=0.24M$

$[H_2O]=\frac{5}{100}=0.05M$

Putting values in expression 1, we get:

$K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3$

Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 24+11 5 10

At eqllm: 35-x 5-x 10+x

$[CH_3CH_2OH]=\frac{10+x}{100}$

$[CH_2=CH_2]=\frac{35-x}{100}$

$[H_2O]=\frac{5-x}{100}$

Putting values of in expression 1, we get:

$8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1$

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.

Moles of ethanol = $(10+x)=10+1.1=11.$

Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.

8 0

3 years ago

Can someone explain me this with answer? I will make brain list

Ostrovityanka [42]

Answer:

Sawdust floats, sand dissolves, and 3rd dissolves

4 0

3 years ago