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3 years ago

A supercell thunderstorm may have a distinctive shape like _____.

1 answer:

8 0

D. Pinwheel probably because think it’s a thunderstorm and supercells mean they are super big thunderstorms so I guess D

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What structure do bacteria cells have

g100num [7]

Answer:

They are clumped together and fluffy looking

Explanation:

I’ve seen what they looked like in a project during highschool we swabbed some books and door handles and saw which had the most bacteria. It was surprisingly the books

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1 year ago

What is mole ? and it's unit

Aloiza [94]

$\\ \sf\longmapsto 1mole\:is\:a\:specific\:amount\:of\:particles\:of\:a\:substance .$

$\\ \sf\longmapsto It\:is\:Represented\:as\:mol$

$\\ \sf\longmapsto 1mol=6.023\times 10^{23}particles$

4 0

2 years ago

Read 2 more answers

Question 33 Saved

Tatiana [17]

Answer:

The third option

Explanation:

If people have found more volcanic rock layers in the past that would mean that volcanic activity was more common in the past.

3 0

2 years ago

What is work and state the law of conservation mass?<br>

Butoxors [25]

Answer:

Work is a force causing the movement or displacement of an object

law of conservation mass:

1. Atoms cannot be created or destroyed in a chemical reaction.

2. Molecules cannot be created or destroyed in a chemical reaction.

3. Compounds cannot be created or destroyed in a chemical reaction.

4. Heat cannot be created or destroyed in a chemical reaction.

7 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

2 years ago