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Juliette [100K]
2 years ago
6

Is flat soda mixture or pure substance?

Chemistry
1 answer:
Nesterboy [21]2 years ago
5 0
The answer should be Pure substance
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Write the skeleton equation: <br> Cobalt and Sulfur react to produce Cobalt (II) Sulfide
lidiya [134]

Answer:

Co + S = Co2S3

Explanation:

pretend = is the arrow

7 0
2 years ago
Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta
scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion=\frac{Volume}{Time}

Effusion rate of oxygen gas after time t = E=\frac{4.64 mL}{t}

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = E'=\frac{3.12 mL}{t}

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{E}{E'}=\sqrt{\frac{M'}{M}}

\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

8 0
3 years ago
Calculate the number of Carbon atoms in 1.50g of C6H12O6
Whitepunk [10]

Answer:

3.01 ·10↑22

Explanation:

First you want to convert the grams of Glucose to moles of Glucose.

\frac{1.5 grams of glucose}{180.15588grams/molesglucose} =.008326 Moles of Glucose

Next find the formula units of glucose.

.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =

5.01 ·10↑21 Formula Units of Glucose

Now multiply the formula units of glucose by the amount of each element in the molecule.

So for Carbon:

6carbon · 5.01 · 10↑21 = 3.01 · 10↑22

(please mark as Brainliest)

4 0
3 years ago
Read 2 more answers
How many numbers of atoms there are in 1.14 mol of (SO3)
Pachacha [2.7K]

called the Avogadro number 

N(A)= 6.02 x 10^23 mol^-1 

1 mole of SO3 will contain 6.02 x 10^23 mol^-1 of SO3 molecules. 

thus, 1.14moles will contain; 

= 1.14mol x [3mol O/1mol SO3] x [6.02 x 10^23

atoms O/1mol O] 

= 2.05884 x 10^24 oxygen atoms 

= 1.14mol x [1mol S/1mol SO3] x [6.02 x 10^23

atoms O/1mol O] 

= 6.8628 x 10^23 sulfur atoms 

hope this helps:-)

8 0
3 years ago
Which statement is true about a catalyst?
mixas84 [53]

Answer:

Option B: It remains unused during the reaction

Explanation:

Just took the test and got it right.

8 0
3 years ago
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