If you burn 59.1 g of hydrogen and produce 528 g of water , how much oxygen is reacted

The water for the pan is not saturated with co2. will the reported percent caco3 in the original sample be too high, too low, or

enot [183]

Co2 will dissolve in water if water is saturated with Co2 first then it will absorb some of CO2 it appear as less is absorb that was actually the case . this would make it appear as present of CACO3 is in original sample is lower than it will be too low

8 0

3 years ago

15 points + brainliest pls help :(

Komok [63]

Answer:

27 g

Explanation:

M(C6H12O6) = 6*12 + 12*1 + 6*16 = 180 g/mol

100 mL = 0.1 L solution

1.5 M = 1.5 mol/L

1.5 mol/L * 0.1 L = 0.15 mol C6H12O6

0.15 mol * 180 g/1 mol = 27 g C6H12O6

6 0

2 years ago

Someone help please , I’ll give brain . Question 10 !

Rudik [331]

Answer:it’s c

Explanation:

I just know it is

4 0

3 years ago

What is the significance of "Er" in the diagram?

Drupady [299]

Answer:

The significance of "Er" in the diagram is :

B.) Threshold energy for reaction

Explanation:

Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .

Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.

Activation Energy = Threshold Energy - Average Kinetic Energy

This means Activation energy decreases on increasing kinetic energy

On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.

Catalyst also reduces the Activation energy.

Er = Threshshold energy for reaction at 30 degree

Ea = Activation Energy

The given figure shows that the threshold energy decreases on increasing the temperature

Only the molecule having energy greater than Er can react to form product

3 0

3 years ago

Read 2 more answers

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

A

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

C

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

2 years ago