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Liula [17]
3 years ago
10

Read ""the Ozone Hole"" and answer the question below list at least three scientific disciplines related to chemistry mentioned

or alluded to in the article
Chemistry
1 answer:
I am Lyosha [343]3 years ago
8 0
Biology, astronomy, and climate science
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How to solve 20.0 mL/ 5.7 g with significant figures and including units
Nutka1998 [239]

Question

How to solve 20.0 mL/ 5.7 g with significant figures and including units?

Answer:

p = 0.25 g/cm3

Hope this Helps!!

xXxAnimexXx

The formula for density is the mass of an object divided by its volume. In equation form, that's d = m/v , where d is the density, m is the mass and v is the volume of the object. The standard units are kg/m³.

7 0
3 years ago
SEP math! Each day, the average person's
Lesechka [4]

Answer:

Quarts of blood = 72000

Quarts of urine = 720

Explanation:

<h3><u>Given</u>;</h3>
  • Each day, the average person's
  • kidneys filter around 200 quarts of blood.
  • About 2 quarts of urine is produced in the
  • process.
<h3><u>To </u><u>Find</u>;</h3>
  • In one year the average person's kidney filter quarts of blood and quarts of urine.

Here, we know that

1 year = 365 days

For blood

200 × 365 = 72000

For urine

2 × 365 = 720

Thus, The average person's kidney filters 72000 quarts of blood and 720 quarts of urine.

5 0
1 year ago
In which way are evaporation and condensation similar?
shepuryov [24]
1. Both evaporation and condensation are steps in the water cycle.

2. Both are affected by atmospheric conditions.

Let me know how you did, good luck! :D
7 0
3 years ago
Read 2 more answers
The combustion of a 0.4255 g sample of a compound containing carbon, hydrogen, oxygen and bromine produces 0.4961 g CO2 and 0.17
IgorC [24]

Answer: The empirical formula is C11HO3Br

Explanation: 1ST SAMPLE;

First, we need to get the number of moles;

Molar mass of C=12, O=16, H=1, Ag =107.9, Br=79.9

0.4961g CO2 x (1 mol CO2) / (44.0g CO2) = 0.0113 mol of Co2

Since one mole of CO2 is made up of 1 mole of C and 2 moles of O, and we have 0.0113 moles of CO2 in our sample, then we know we have 0.0113 moles of C in the sample. Now, we need to know the mass of C we have in the sample:

(0.0113 mol C) (12.011g C/1 mol C) = 0.136g C

Now we follow the same pattern above and do it for the hydrogen:

0.1777g H2O x (1 mol H2O) / (18g H2O) = 0.01 mol

Now, for the mass of H:

(0.01 mol H) (1g H/1 mol H) = 0.01g H

But this is for 1 mole of H. Whereas, H has 2 moles from the question, therefore it's equal to 0.01 x 2 = 0.02g H

Since we combusted 0.4255g of the sample, the missing mass will be from the bromine and oxygen since we have gotten masses of Carbon and Hydrogen.

Therefore missing mass = 0.4255 - (0.136+0.02) = 0.2695 of bromine and oxygen

2ND SAMPLE:

First, we need to get the number of moles;

0.1894 g AgBr x (1 mol AgBr) / (107.9 + 79.9g AgBr) = 0.001 mol of AgBr

Since AgBr is made up of 1 mole of Ag and 1 mole of Br each,

We can say that there's 0.001 mole of Br in this second sample.

Now looking at the first and second samples, lets set up a proportion to know the number of moles of Br in the first sample.

If 0.1523g of the second sample produced 0.001 mol of Br, therefore 0.4255g in the first sample will produce: (0.4255g x 0.001)/0.1523 = 0.0028mol of Br

Therefore the mass of Br in the first sample is: (0.0028 mol C) (79.9g Br/1 mol C) = 0.22372g Br

From the first sample, we saw that the sum of Br and Oxygen equals 0.2695

Therefore,the mass of oxygen is: 0.2695 - 0.22372 = 0.04578g of oxygen

Therefore to find number of moles of oxygen;

(0.04578g O x (1 mol O) / (16.0g O) = 0.0029 mol of oxygen

Overall, we have; C=0.0113 moles ;H=0.001 moles; Br= 0.0028moles and O= 0.0029

The smallest is 0.001. So to simplify this for the empirical formula, we divide each by 0.001 to get approximately C= 11, H=1, Br=3, O=3

Therefore the empirical formula is C11HO3Br

5 0
3 years ago
Question 3 of 25
Lisa [10]

Answer:

I thinks it Br hope this helps

4 0
2 years ago
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