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3 years ago

7

A brownish haze that is a mixture of ozone and other chemicals, formed when pollutants react with each other in the presence pf

sunlight

1 answer:

soldi70 [24.7K]3 years ago

8 0

The haze is called smog, hope this helps, pls mark brainliest!

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Use the Gizmo to estimate the age of each of the objects below. For these questions, each second in the Gizmo represents 1,000 y

ziro4ka [17]

Answer:

Check the explanation

Explanation:

AT = A0 e(-T/H)

... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.

Substituting what we know, we get...

0.71 = (1) e(-T/5730)

Solve for T...

loge(0.71) = -T/5730

T = -loge(0.71)(5730)

T = 1962 (conservatively rounded, T = 2000)

similarly for all

for aboriginal charcoal

0.28 = (1) e(-T/5730)

Solve for T...

loge(0.28) = -T/5730

T = -loge(0.28)(5730)

T = 7294 (conservatively rounded, T = 7000)

for mayan headdress

0.89 = (1) e(-T/5730)

Solve for T...

loge(0.89) = -T/5730

T = -loge(0.89)(5730)

T = 667 (conservatively rounded, T = 700)

for neanderthal

0.05 = (1) e(-T/5730)

Solve for T...

loge(0.05) = -T/5730

T = -loge(0.05)(5730)

T = 17165 (conservatively rounded, T = 17000)

7 0

3 years ago

Balance the following chemical equation by providing the correct coefficients fe+h2so4 fe(so4)3 + h2

Dmitriy789 [7]

Answer:

2Fe + 3H2SO4 + Fe2(SO4)3+ 3H2

Explanation:

1. Fe (SO4) 3 is an incorrectly written formula because iron is trivalent as we can see by this three ahead of SO4. SO4 is divalent always.

2. since (SO4) is 3, this three shows us that there must be 3 in the reactants as well.

so now there is 3H2SO4

3. Since we have added 3 to one hydrogen we must add another. So now it's 3H2

4. and finally iron. In Fe2 (SO4) 3 we see this 2 in front of Fe which means it goes 2Fe.

3 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.

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3 years ago

What is the charge of oxygen in Na2O?

Nuetrik [128]

In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+

3 0

2 years ago

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Given an unknown mixture consisting of two substances, explain how a scientist could use lab techniques to determine whether the

Katyanochek1 [597]

<span>Pass the mixture through filter paper. The large particles in the suspension will filter out. to tell the difference between a solution and a colloid, shine a beam of light through the mixture, if it reflects then it is a colloid, if it doesn't then it is a solution</span>

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3 years ago