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Anni [7]
3 years ago
12

Pollution comes from many small sources.

Chemistry
1 answer:
Digiron [165]3 years ago
5 0

Answer: True

Hope this helps

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Why is earth's surface warmer in the northern hemisphere when it is summer there
RideAnS [48]
 <span>By definition, summer is the portion of the year in which the hemisphere is tilted toward the sun, so that sunlight strikes the surface more directly. When it is summer in the northern hemisphere, it is winter in the southern hemisphere, and vice versa</span>
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3 years ago
Read 2 more answers
A slice of Swiss cheese contains 47 mg of sodium. (a) What is this mass in grams? .047 (b) What is this mass in ounces? (16 oz =
MAVERICK [17]

Answer:

(a) 0.047 g (b) 0.0016 oz (c) 0.0001 lb

Explanation:

The given mass of the sodium in the slice = 47 mg

(a) Mass has to be calculated in grams

The conversion of mg to g is shown below as:

1 mg = 10⁻³ g

So,

<u>Mass of sodium = 47 × 10⁻³ g = 0.047 g</u>

(b) Mass has to be calculated in ounces

The conversion of ounces to g is shown below as:

453.6 g = 16 oz

Or,

1 g = 16 / 453.6 oz

So,

<u>Mass of sodium = (0.047 × 16) / 453.6 oz = 0.0016 oz</u>

(c) Mass has to be calculated in pounds

The conversion of pounds to g is shown below as:

1 lb = 453.6 g

Or,

1 g = 1/ 453.6 lb

So,

<u>Mass of sodium = (0.047 × 1) / 453.6 oz = 0.0001 lb</u>

5 0
2 years ago
What is the mass in grams of 0.7350 moles of sodium?
Alik [6]

Answer:

Explanation:

1 mol of sodium = 23 grams (use the number on your periodic table).

0.7350 mol sodium = x

Cross multiply

1*x = 0.7350 * 23

x = 16.905

You will get slightly less than this, depending on your periodic table. But the method will be the same.

4 0
2 years ago
I cant remember my mass, grams, and volume. and how to divide density
Bas_tet [7]
Mass divide by volume

M
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V
7 0
2 years ago
What is the limiting reactant for the following balanced equation when 9 moles of AlF3 are mixed with 12 miles of O2?
tamaranim1 [39]
<h2>Answer:AlF_{3} </h2>

Explanation:

The chemical equation of the reaction that occurs when AlF_{3} reacts with O_{2} is

4AlF_{3}+3O_{2}→2Al_{2}O_{3}+6F_{2}

4 moles of AlF_{3} requires 3 moles of O_{2}.

1 mole of AlF_{3} requires \frac{3}{4} moles of O_{2}.

Given that we have 9 moles of AlF_{3}.

9 moles of AlF_{3} requires \frac{3}{4}\times 9=6.75 moles of O_{2}.

But we have 12 moles of O_{2}.

So,AlF_{3}  will be consumed first.

So,AlF_{3}  is the limiting reagent.

3 0
3 years ago
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