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Nataly_w [17]
3 years ago
10

Which is a combination of pollutants that mix with water vapor in a city or near large highways and factories?

Chemistry
1 answer:
Valentin [98]3 years ago
8 0
Me mememememememememememememememememeememememeemememememememe
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Discuss FOUR ways in which one's acceptance of responsibility can influence
andrew11 [14]

Answer:

being responsible is owing up to the possible

consequences of your decision whether it was right or wrong.

communication is a two way process in which there is a messenger and receiver

one of the ways by which a person could be responsible is by saying 'I'

in all the things he would say, instead of 'we'

for example , " We don't like the way you acted a while ago " , change it to i don't like tye way you acted a while ago.

here u r establishing ownership on the things you want to say to the person

8 0
3 years ago
If 36 000 kg of full cream milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream
Yuri [45]

Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

-XgA = percentage of fat in A = 0.04

We define output variables.

- skim milk (B)

-creme (C)

-XgB = fat percentage at B = 0.0045

-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

Input = Output

Balance sheet

1) A = B + C

Fat balance

2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values ​​of the outflows.

C=478.11kg/h

B=5522.33kg/h

7 0
3 years ago
If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?
bija089 [108]
<h3>Molar mass:-</h3>

\\ \sf\longmapsto HC_2H_3O_2

\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)

\\ \sf\longmapsto 1u+24u+3u+48u

\\ \sf\longmapsto 28u+48u

\\ \sf\longmapsto 76u

\\ \sf\longmapsto 76g/mol

  • No of moles=0.2mol
  • Given mass=?

\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}

\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}

\\ \sf\longmapsto Given\:Mass=0.2\times 76

\\ \sf\longmapsto Given\:Mass=1.52g

7 0
3 years ago
What do we call genes
qwelly [4]

Hi there so i need one more brainliest mark so if i get it right can i get brainliest please  :)

A gene is the basic physical and functional unit of heredity. Genes are made up of DNA. Some genes act as instructions to make molecules called proteins. However, many genes do not code for proteins. In humans, genes vary in size from a few hundred DNA bases to more than 2 million bases

4 0
2 years ago
How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1
valentinak56 [21]
<span>4 Al + 3 O2 → 2 Al2O3 

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al 
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2 

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess. 

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) = 
10.1 g O2 left over</span><span>
</span>
7 0
3 years ago
Read 2 more answers
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