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3 years ago

Please help me I'm really confused on how to do this. Can someone break it down for me? (stoichiometry)

Chemistry

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

so with every stoichiometry problem with a mass it will make it so you can do the conversion factor with reactants or products.

if you dont understand unit conversions try to study how to set it up. anyways

a.) C12H22O11 has a mass of 342.01 Grams per mole

divide 1.202 G by 342.01 G to get 0.004 miles

b.) you're just taking the AMU of each element in the chemical multiply it by how many there is of it in the chemical, then divide it by the mass of a mole of the chemical.

c.) you take your answers of part b and multiply them by Avogadro's number

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A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

3 years ago

PLEASE HELP WITH ENVI SCI QUESTION

Vladimir79 [104]

Different types of acid deposition are

sulphur dioxide
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3 years ago

4. If the DNA nitrogen bases were TACCGGAT, how would the other half of

Illusion [34]

Answer:

ATGGCCTA

Explanation:

For this we have to keep in mind that we have a specific relationship between the nitrogen bases:

-) When we have a T (thymine) we will have a bond with A (adenine) and viceversa.

-) When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa.

Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:

TACCGGAT

ATGGCCTA

I hope it helps!

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The answer is D. Heterogeneous mixture

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