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3 years ago

Beta particles cannot penetrate very far into solids because they _____.

2 answers:

6 0

Answer: Beta particles cannot penetrate very far into solids because they <span><u>have no mass.

</u>Reason<u>:
</u>Penetrating power of particle depending upon the velocity and mass of particle. Greater the mass and velocity. Higher will be the penetrating power. In case of Beta particles, they are electrons. Mass of an electron is 9.1 X </span> $10^{-31}$ Kg. Due to such negligible mass, they have very low penetrating power. <span /><span><u>
</u></span>

IRINA_888 [86]3 years ago

3 0

Answer:have no mass

Explanation:

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30.0L of helium gas into moles of helium gas.

zepelin [54]

<h3>Answer:</h3>

1.25 moles (R.T.P.) or 1.34 moles (S.T.P.)

<h3>Explanation:</h3>

1 mole of a gas occupies a volume of 24 liters at room temperature and pressure (R.T.P.)
On the other hand, 1 mole of a gas will occupy 22.4 Liters at standard temperature and pressure (S.T.P.)

Therefore, at R.T.P.

30.0 Liters will be equivalent to;

= 30.0 L ÷ 24 L

= 1.25 moles

At S.T.P

30.0 Liters will be equivalent to;

= 30.0 L ÷ 22.4 L

= 1.34 moles

Thus, 30.0 L of helium gas are equivalent to 1.25 moles of He at R.T.P. and 1.34 moles at S.T.P.

3 0

3 years ago

Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen

Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen= $n_1$

Moles of acetylene = $n_2$

$\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}$

Volume of large tank with oxygen gas, $V_1 = 5.00 L$

Pressure of the oxygen gas inside the tank = $P_1=135 atm$

$RT=\frac{P_1V_1}{n_1}$ ..[1]

Volume of small tank with acetylene gas , $V_2= 3.50 L$

Pressure of the acetylene gas inside the tank = $P_2=?$

$RT=\frac{P_2V_1}{n_2}$ ..[2]

Considering both the gases having same temperature T, [1]=[2]

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

$P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}$

$=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm$

77.14 atm of pressure should be of an acetylene in the tank.

8 0

3 years ago

Which chemical element has the shortest name.

Westkost [7]

Tin is the shortest chemical element on the Periodic Table.

8 0

3 years ago

Read 2 more answers

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo

frozen [14]

Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

$W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})$

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

$W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J$

Stape 2: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

$W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J$

The total work is the sum of the two steps

$W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J$

4 0

3 years ago

Number of electrons present in 10th shell of<br> an atom

stellarik [79]

Answer:

there will be 8 electrons

3 0

3 years ago