Which of the following methods can be used to identify the suspected use of heat lamps in a house without entering the house?

Of the choices below, which is true for the relationship shown? it is ka for the acid h3p2o72â. it is kb for the acid h3p2o72â.

Semenov [28]

A For PLATO. just did it

3 0

2 years ago

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A woman weighing 450 Newtons sits on the floor. She exerts a force on the floor of

ANEK [815]

Answer: let me figure it out rlly quick

Explanation:

4 0

2 years ago

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

Q4

Varvara68 [4.7K]

Answer:

Option C.

Ilegal dumping of waste

Explanation:

This is because non point source of pollution refers to source of pollution that are many and not directly one which is illegal or does not meet the legal term. This type of pollution does not have a point source, it has many sources and this type of pollution is cause by rainfall or precipitation. Where when the rain fall, it wash away the waste through to water bodies, causing pollution and endangering water bodies.

3 0

3 years ago

Write the mole ratios of (2ca+o2➡2cao)

tresset_1 [31]

N(Ca)/2 = n(O)/1 = n(CaO)/2

The calcium and the Calcium Oxide are divided by 2 because of their coefficients
there is no number in front of the oxygen so it is over one.
Hope this helped!!

6 0

3 years ago

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