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True [87]
3 years ago
13

Sulfur dioxide and nitrogen dioxide are considered to

Chemistry
1 answer:
quester [9]3 years ago
8 0

a)CO is the main air pollutant (b) All pollutants are not wastes (c) Water is polluted by dissolved oxygen (d) Lichens are pollution indicatorsnation:

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How many moles are in 128.9 grams of Cr2(SO3)2?
Dominik [7]

Answer: 0.5 moles

Explanation:

Cr2(SO3)2 is the chemical formula for chromium sulphate.

Given that,

Amount of moles of Cr2(SO3)2 (n) = ?

Mass of Cr2(SO3)2 in grams = 128.9g

For molar mass of Cr2(SO3)2, use the atomic masses:

Chromium, Cr = 52g;

Sulphur, S = 32g;

Oxygen, O = 16g

Cr2(SO3)2 =

(52g x 2) + [(32g + 16g x 3) x 2]

= 104g + [(32g + 48g) x 2]

= 104g + [80g x 2]

= 104g + 160g

= 264g/mol

Since, n = mass in grams / molar mass

n = 128.9g / 264g/mol

n = 0.488 mole [Round the value of n to the nearest tenth which is 0.5

Thus, there are 0.5 moles in 128.9 grams of Cr2(SO3)2

7 0
3 years ago
Please help
KonstantinChe [14]

Answer:  There are 0.5 grams of barium sulfate are present in 250 of 2.0 M BaSO_{4} solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given BaSO_{4} solution is as follows.

Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M BaSO_{4} solution.

7 0
3 years ago
Identify the term that matches each definition.
Nikolay [14]

Answer:

a. Volatile.

b. Air foil.

c. Sash.

d. Work surface.

Explanation:

In science, matter can be defined as anything that has mass and occupies space. Any physical object that is found on earth is typically composed of matter. Matter are known to be made up of atoms and as a result has the property of existing in states.

Generally, matter exists in three (3) distinct or classical phases and these are;

I. Solid.

II. Liquid.

III. Gas.

Matching the terms with their respective definition, we have;

a. Volatile: A characteristic that describes substances that evaporate readily, producing large amounts of vapors.

b. Air foil: the front vent of a fume hood, which helps maintain proper air circulation.

c. Sash: the glass panel in front of the fume hood that shields the user from fumes and other hazards.

d. Work surface: the horizontal, flat area of a fume hood upon which experiments are carried out.

7 0
3 years ago
PLEASE HELP WITH THESE!! I’m stuck
ludmilkaskok [199]
3. B
4. A
5.A
6. A (i think)
Hope this helps
4 0
3 years ago
Calculate the theoretical yield of 1-bromobutane; base your calculations on using 1.0 g of 1-butanol (as the limiting reagent).
Tomtit [17]

C₄H₉OH + HBr = C₄H₉Br + H2O

Δmole of alcohol gives 1 mole of bromobutanol

HBr is in excess, so the yield of the product is limited by the alcohol

Wt. of 1 butanol = 18

Molar mass of the butanol = 74.12 g/mole

Moles of the alcohol = 1/74.12 = 0.01349 moles

So, moles of bromobutane = 0.01349 moles

Molar mass of C₄H₉Br = 137.018 g/moles

So, theoretical mass of bromobutane is = 0.01349 × 137.0.18

= 1.85 g


6 0
4 years ago
Read 2 more answers
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