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andreev551 [17]
3 years ago
5

A graduated cylinder has 35.0ml of water in it before an object is dropped inside. The new volume is 38.5ml. What is the volume

of the object in cm3
Chemistry
1 answer:
natima [27]3 years ago
4 0
1 mL = 1 cm³

<span>Displaced volume :
</span>
Vf - Vi

38.5  - 35.0 => 3.5 cm³

hope this helps!


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There are 43.2 g of carbon to 14.4 g of hydrogen in a sample of methane (CH4). What percent of 37.8 g of methane is carbon?
EastWind [94]

Using the law of constant proportions  which says that within the same compound, elements exist in fixed ratios. 

Therefore; we can use the ratio of total mass to the mass of carbon, to determine the amount of carbon in another sample.

Mass C / Mass CH4 = Mass C / Mass CH4

43.2 g / 57.6 g = Mass C / 37.8 g

Mass C = 37.8 g × 43.2 g / 57.6 g 

              = 28.35 g

Hence; the percentage of carbon will be; 

=(28.35/ 37.8 )× 100%

= 75 % 

Thus; 75% of 37.8 g of methane is carbon

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Is it true that Organisms need water, vitamins and minerals. But they do not need salt to function.
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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
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