Can u please help me with these 2 answers!

You carefully weigh out 14.00 g of CaCO3 powder and add it to 56.70 g of HCl solution. You notice bubbles as a reaction takes pl

zepelin [54]

For the answer to the question above, let us assume that all Co2 is given off and this are the mass that is lost:

mass at start = 14.00 = 56.70 = 70.70g
mass at end = 64.96g 
mass lost = 5.74g 

I hope my answer helped you. Have a nice day!

7 0

4 years ago

Read 2 more answers

Name a sensory organ, and briefly explain its function.

Sveta_85 [38]

Your nose is a sensory organ and it’s function is to be able to smell chemicals, it cleans the air you breathe, it regulates the temp of the air you breathe, and it’s the main route for your breathing.

6 0

3 years ago

Heavier compounds tend to have higher boiling points. However, sulphur dichloride (Mr= 103.1 gmol-1) and sulphur hexafluoride (M

choli [55]

Hm im not sure maybe c

7 0

3 years ago

Measurements show that unknown compound has the following composition: element mass 62.1 % carbon, 10.5 % hydrogen and 27.6 % ox

barxatty [35]

Answer:

C3H6O

Explanation:

The percentage composition of the elements in the compound are given as follows:

62.1 % carbon = 62.1g of C

10.5 % hydrogen = 10.5g of H

27.6 % oxygen = 27.6g of O

Next, we convert each mass to mole by dividing by their molar/atomic mass

C = 62.1/12 = 5.175mol

H = 10.5/1 = 10.5mol

O = 27.6/16 = 1.725mol

Next, we divide each mole value by the smallest mole value (1.725)

C = 5.175mol ÷ 1.725 = 3

H = 10.5mol ÷ 1.725 = 6.086

O = 1.725mol ÷ 1.725 = 1

The empirical ratio approximately of C:H:O is 3:6:1, hence, the empirical formula is C3H6O

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