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Maksim231197 [3]
3 years ago
6

Please help me i don’t understand.

Chemistry
1 answer:
vekshin13 years ago
8 0

Answer:

Your answer is

p = m  \div v \\  =  \: 38 \: gram \:  \div 2 \: cm  \\   = 19 \: gram \:  = 19000gm

Hope it helped

You might be interested in
what is the ratio of the rate of effusion of helium (atomic mass 4.00 amu) to that of oxygen gas (molecular mass 32.0 amu)?
nignag [31]

Answer:

3 : 1

Explanation:

Let the rate of He be R1

Molar Mass of He (M1) = 4g/mol

Let the rate of O2 be R2

Molar Mass of O2 (M2) = 32g/mol

Recall:

R1/R2 = √(M2/M1)

R1/R2 = √(32/4)

R1/R2 = √8

R1/R2 = 3

The ratio of rate of effusion of Helium to oxygen is 3 : 1

8 0
2 years ago
I need help solving this!
zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

5 0
3 years ago
(Only answer if you’re for certain) Which type of molecule is shown below?
polet [3.4K]

i believe B is the answer

4 0
3 years ago
Read 2 more answers
Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low
LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

4 0
3 years ago
A base
hichkok12 [17]
B, turns red litmus paper to blue
5 0
3 years ago
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