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3 years ago

12

A rectangle solid of unknown density is 5 m long two meters high and 4 meters wide. The mass of this solid is 300 grams. Find th

e volume and density

1 answer:

aksik [14]3 years ago

6 0

D= M/V so, D=300/40 wich is 7.5 so D= 7.5g/m3

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Which of the following is an

eimsori [14]

Answer:

2 and 4

Explanation:

The rest of the changes are chemical. 1 has a chemical reaction happen which makes light sticks glow. 3 is because browning the meat actually causes some new compounds to form and cause caramelization.

For 2, the oxygen is simply heating up and expanding which pops the balloon. 4 is just a phase change of water vapor to liquid water.

5 0

3 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

Where more of hydrogen are contained in 90 g of water or in 80 g of methane

Lynna [10]

Methane.

Water - H2O
Methane - CH4

Methane has 2 more hydrogens than water.

7 0

3 years ago

The balanced equation for combustion in an acetylene torch is shown below: 2C2H2 + 5O2 → 4CO2 + 2H2O The acetylene tank contains

andriy [413]

Answer: 67.2 moles

Explanation: $2C_2H5+5O_2\rightarrow 4CO_2+2H_2O$

According to the given balanced equation, 2 moles of acetylene $C_2H_2$ combine with 5 moles of oxygen $O_2$ to produce 4 moles of carbon dioxide $CO_2$ .

Thus if 2 moles of acetylene $C_2H_2$ combine with = 5 moles of oxygen $O_2$

35 moles of acetylene $C_2H_2$ combine with= $\frac{5}{2}\times {35}=87.5$ moles of oxygen $O_2$

But as only 84 moles of oxygen are available, acetylene is not a limiting reagent.

5 moles of oxygen $O_2$ reacts with = 2 moles of acetylene $C_2H_2$

84 moles of oxygen $O_2$ reacts with= $\frac{2}{5}\times {84}=33.6$ moles of acetylene $C_2H_2$

Thus Oxygen is the limiting reagent as it limits the formation of products. Acetylene is excess reagent as it is present in excess.

2 moles of acetylene $C_2H_2$ produce= 4 moles of carbon dioxide $CO_2$ .

33.6 moles of acetylene $C_2H_2$ produce= $\frac{4}{2}\times {33.6}=67.2$ moles of of carbon dioxide $CO_2$ .

8 0

3 years ago

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From his experiment j.j. Thomson concluded that

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Particles as small as atoms exist.

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3 years ago