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puteri [66]
3 years ago
13

How many atoms of hydrogen are in 1 mole of Dimethylnitrosamine (CH3)2N20?

Chemistry
1 answer:
Alex Ar [27]3 years ago
4 0

there are 6 total Hydrogen atoms in Dimethylnitrosamine

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How many moles are in 135g of Teflon?
Oliga [24]

1.34 moles are in 135g of Teflon.

Explanation:

Molecular formula of teflon is (C2H4)n

mass of teflon = 135 grams

atomic mass of teflon = 100.01 gram/mole

number of moles of teflon = ?

Formula used to calculate number of moles in a substance is given as:

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the above equation:

number of moles = \frac{135}{100.01}

number of moles = 1.34 moles

Teflon is polymer which is used for making non-stick coating, in 135 grams of teflon there are 1.34 moles in it.

5 0
3 years ago
Help PLEASEEE !!!!!! will give brainliest !
BabaBlast [244]

Answer:

6. O₂ + Cu —> CuO

7. H₂ + Fe₂O₃ —> H₂O + Fe

8. O₂ + H₂ — > H₂O

9. H₂S + NaOH —> Na₂S + H₂O

10. Al + HCl —> H₂ + AlCl₃

Explanation:

6. Oxygen gas react with solid copper metal to form copper(II) oxide

Oxygen gas => O₂

Copper => Cu

copper(II) oxide => CuO

The equation is:

O₂ + Cu —> CuO

7. hydrogen gas and iron(III) oxide powder react to form liquid water and solid iron power

hydrogen gas => H₂

Iron(III) oxide => Fe₂O₃

Water => H₂O

Iron => Fe

The equation is:

H₂ + Fe₂O₃ —> H₂O + Fe

8. Oxygen gas react with hydrogen gas to form liquid water

Oxygen gas => O₂

hydrogen gas => H₂

Water => H₂O

The equation is:

O₂ + H₂ — > H₂O

9. Hydrogen sulphide gas is bubbled through a sodium hydroxide solution to produce sodium sulphide and liquid water

hydrogen sulphide => H₂S

sodium hydroxide => NaOH

Sodium sulphide => Na₂S

Water => H₂O

The equation is:

H₂S + NaOH —> Na₂S + H₂O

10. Hydrogen gas and aluminum chloride solutions are produced when solid aluminum react with hydrochloric acid

Aluminum => Al

Hydrochloric acid => HCl

hydrogen gas => H₂

Aluminum chloride => AlCl₃

The equation is:

Al + HCl —> H₂ + AlCl₃

5 0
3 years ago
Which term is described as a long, narrow depression in the ocean floor?
ahrayia [7]
The term which is described as a long, narrow depression in the ocean floor would be ocean trench. They <span>are hemispheric-scale long but narrow topographic depressions of the sea floor. They are also the deepest parts of the ocean floor. Hope this answers the question.</span>
8 0
3 years ago
Read 2 more answers
28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cir
Damm [24]
<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

         = 0.0063 mole

<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

                             = 1.21 g citric acid

<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

   = (0.0050 × 0.25)÷ 0.30

   = 0.004167 L or

   = 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

8 0
3 years ago
What is the density of and object that is 100 grams and has a volume of 5 cubic centimeters?
lesya [120]

Explanation:

Density of object = gram/volume

= 100 g/ 5 cm³

= 20 g/cm³

5 0
3 years ago
Read 2 more answers
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