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PtichkaEL [24]
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
Chemistry
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
iragen [17]
Particles as small as atoms exist.
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3 years ago
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