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Kamila [148]
3 years ago
7

Need help answering the question at the bottom of the text

Chemistry
1 answer:
Rasek [7]3 years ago
7 0

Answer:

You Need to put in the picture

Explanation:

I cant see it

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How do you calculate the number of moles in CO2
gogolik [260]

Answer:

The number of molecules in a mole (known as Avogadro's constant) is defined such that the mass of one mole of a substance, expressed in grams, is equal to the mean molecular mass of the substance. The molecular mass of CO2 = 12+2x16 = 44, so the mass of a mole of CO2 is approximalty 44 grams

Explanation:

3 0
3 years ago
How could you verify that you produced carbon dioxide in your combustion reaction? 2. what indication did you have that nh3 was
DanielleElmas [232]

1) Carbon dioxide is a gas, so when CO_{2} is evolved in the reaction, it appears as bubbles. The gas released extinguishes the fire and it can turn lime water milky.

Ca(OH)_{2} (aq) + CO_{2}(g) ----> CaCO_{3}(s) +H_{2}O(l)

2) When NH_{3} is released in a decomposition reaction we can identify by the strong pungent smell of the gas released.

3) Saturated citric acid can cause corrosion of the metal layers present in the pipes. So, before draining out any acid it is neutralized so that the pipes and other plumbing works do not get damaged leading to leaks in the drainage system.

5 0
4 years ago
How many hydrogen atoms are there in the molecule H2SO?
sergiy2304 [10]

Answer:

D:2

Explanation:

H is hydrogen, and the subscript represents the amount of atoms it has so H2SO is two hydrogens

3 0
3 years ago
Read 2 more answers
Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many
Flauer [41]

Answer : The amount of formaldehyde permissible are, 5.4\times 10^{-6}g

Explanation : Given,

Density of air = 1.2kg/m^3=1.2g/L     (1kg/m^3=1g/L)

First we have to calculate the mass of air.

\text{Mass of air}=\text{Density of air}\times \text{Volume of air}

\text{Mass of air}=1.2g/L\times 6.0L

\text{Mass of air}=7.2g

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 5.4\times 10^{-6}g

Thus, the amount of formaldehyde permissible are, 5.4\times 10^{-6}g

8 0
3 years ago
Hydrogen sulfide is composed of two elements: hydrogen and sulfur. in an experiment, 6.500 g of hydrogen sulfide is fully decomp
LiRa [457]
Hydrogen sulfide = hidrogen + sulfur

       6.500 g                      

a)                             0.384 g   +    x

=> 6.500 = 0.384 + x => x = 6.500 - 0.384 = 6.116 g

Answer: 6.116 g of sulfur must be obtained

b) this experiment demonstrate the conservation of mass.

c) Dalton's atomic model states that the atoms cannot be created, split or be destroyed, and so in a chemical reaction the atoms rearrange but the number of each type of atoms remain constant, so the mass of each type of atoms and the total mass remain constant.
8 0
4 years ago
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