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IRISSAK [1]
2 years ago
9

How many grams are there in 7.5 X10^23 molecules of H2SO4?

Chemistry
2 answers:
mash [69]2 years ago
6 0
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

 </span><span>7.5 X10^23 molecules of H2SO4 ( 1 mol / 6.022x10^23 molecules ) (98.08 g / 1 mol ) = 122.15 grams H2SO4</span>
mel-nik [20]2 years ago
4 0

Answer : The mass of H_2SO_4 are, 117.6 grams.

Explanation :

First we have to calculate the moles of H_2SO_4.

As, 6.022\times 10^{23} molecules of H_2SO_4 present in 1 mole of H_2SO_4

So, 7.5\times 10^{23} molecules of H_2SO_4 present in \frac{7.5\times 10^{23}}{6.022\times 10^{23}}=1.2 mole of H_2SO_4

Now we have to calculate the mass of H_2SO_4.

\text{Mass of }H_2SO_4=\text{Moles of }H_2SO_4\times \text{Molar mass of }H_2SO_4

Molar mass of H_2SO_4 = 98 g/mole

\text{Mass of }H_2SO_4=1.2mole\times 98g/mole=117.6g

Therefore, the mass of H_2SO_4 are 117.6 grams.

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5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperatu
kati45 [8]

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<h3>P = 20atm</h3>

<em />

7 0
1 year ago
Hydrogen peroxide, H2O2, is common in many households and is used as a bleaching agent. It usually comes in a dark, opaque bottl
sergij07 [2.7K]

Answer:

Hydrogen peroxide should be stored in

1) a cool environment

2) with amber bottles away from sunlight

3) with little drops of sodium phosphate

Explanation:

It has been confirmed that heat and light aids in the decomposition of hydrogen peroxide according to the equation; 2H2O2→2 H2O + O2.

This means that hydrogen peroxide must be stored in a cool place. This will reduce its rate of decomposition. Secondly, it should be stored in amber bottles away from light since light also aids in its decomposition.

Thirdly, drops of sodium phosphate may be added to prevent its catalytic decomposition during storage.

8 0
2 years ago
Calculate the ratio of effusion rates of cl2 to f2 .
Lelechka [254]
<span>Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)</span>
4 0
2 years ago
Which factor explains why coal dust in an enclosed space is more explosive than coal dust blown outdoors into an open space ? A.
ZanzabumX [31]

The correct option is B.

Coal dust refers to the powered form of coal. Because of the high surface area of coal dust it is highly prone to dust explosion, which involves rapid combustion of fine particles that are suspended in the air; this usually occur in an enclosed place. Coal dust in an enclosed place is more explosive than coal dust that is blown outdoor in an open space because the coal dust in an enclosed place is more concentrated due to restricted space, thus it is more liable to explosion.

8 0
2 years ago
Read 2 more answers
A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a SoaveRedlich-Kwong gas and with
Setler79 [48]

Explanation:

The given data is as follows.

    Moles of propylene = 100 moles,    T_{i} = 300 K

    T_{f} = 800 K,    V_{i} = 2 m^{3}

    V_{f} = 0.02 m^{3},   C_{p} of propylene = 100 J/mol

Now, we assume the following assumptions:

Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.

            W = mC_{p} \Delta T

     100 moles \times 100 J/mol K (800 - 300) K

                 = 5 \times 10^{6} J

                 = 5 MJ

Thus, we can conclude that a minimum of 5 MJ work is required without any friction.

3 0
2 years ago
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