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Zolol [24]
3 years ago
5

The ____ formula is the combination of chemical symbols and numbers that represent the elements and number of atoms within a com

pound.
Chemistry
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

Chemical

Explanation:

The Chemical formula is the combination of chemical symbols and numbers that represent the elements and number of atoms within a compound.

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Which unit measures mass?
Artyom0805 [142]

kilograms are the unit that measures mass

4 0
3 years ago
Explain acid rain and it effects to the environment​
PSYCHO15rus [73]

Answer:Ian is looking at cells using a microscope. He sees a nucleus and a large vacuole in the central area of a cell. What type of cell is he most likely looking at?

Explanation:

Ian is looking at cells using a microscope. He sees a nucleus and a large vacuole in the central area of a cell. What type of cell is he most likely looking at?

4 0
3 years ago
Calculate the molecular mass of carbon monoxide (CO) by adding the atomic mass of carbon (C) and the atomic mass of oxygen (O):
Bond [772]

 The   molecular   mass  of carbon monoxide (CO)  is  28 g/mol


   <u><em>calculation</em></u>

The  molecular  mass  of CO =  atomic mass  of C + atomic  mass  of O

From  periodic table the atomic  mass  of  C = 12 g/mol and for O = 16 g/mol


Therefore the molecular  mass   of Co = 12 g/mol + 16 g/mol = 28 g/mol

6 0
3 years ago
If 31.6 g of KMnO4 is dissolved in enough water to give 160 mL of solution, what is the molarity?
Zina [86]

Answer:

A. 1.25M

B. 19.98g

Explanation:

A. Data obtained from the question include the following:

Mass of KMnO4 = 31.6 g

Volume = 160 mL

Molarity =..?

We'll begin by calculating the number of mole KMnO4 in the solution. This is can be obtained as follow:

Mass of KMnO4 = 31.6 g

Molar mass of KMnO4 = 39 + 55 + (16x4) = 158g/mol

Number of mole of KMnO4 =..?

Mole = mass /Molar mass

Number of mole of KMnO4 = 31.6/158 = 0.2 mole

Now, we can obtain the molarity of the solution as follow:

Volume = 160 mL = 160/1000 = 0.16L

Mole of KMnO4 = 0.2 mole

Molarity = mole /Volume

Molarity = 0.2/0.16 = 1.25M

B. Data obtained from the question include the following:

Volume = 300mL

Molarity = 0.74 M

Mass of H2C2O4 =..?

First, we shall determine the number of mole H2C2O4. This is illustrated below:

Volume = 300mL = 300/1000 = 0.3L

Molarity = 0.74 M

Mole of H2C2O4 =?

Mole = Molarity x Volume

Mole of H2C2O4 = 0.74 x 0.3

Mole of H2C2O4 = 0.222 mole

Now, we can easily find the mass of H2C2O4 by converting 0.222 mole to grams as shown below:

Number of mole of H2C2O4 = 0.222 mole

Molar mass of H2C2O4 = (2x1) + (12x2) + (16x4) = 2 + 24 + 64 = 90g/mol

Mass of H2C2O4 =..?

Mass = mole x molar mass

Mass of H2C2O4 = 0.222 x 90

Mass of H2C2O4 = 19.98g

5 0
3 years ago
How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The
bezimeni [28]

Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

Explanation:

The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz =  100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

5 0
3 years ago
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