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icang [17]
3 years ago
11

For which of the following materials is necessary to stop a beta particle?

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0
Hello.

The answer would be <span>A. Three feet of concrete.

Have a nice day.</span>
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Hydrogen bond is an example for? (PLEASE HELP!)
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Answer:

b)

Explanation:

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Animals can change the energy in food into energy the animals can use. What type of energy is in the food before the animals cha
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Perform each of the following unit conversions using the conversion factors given below: 1 atm = 760 mmHg = 101.325 kPa (Round a
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4 years ago
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(4x2 + 5x) - (7x - 3x2 + 1)​
Tanya [424]

Answer:

7x2−2x−1

Explanation:

do u need the explanation if so ill add it

8 0
3 years ago
C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????
Yanka [14]

Answer:

Rate constant at 725 K is 5.2\times 10^{-4}s^{-1}

Explanation:

According to Arrhenius equation for a reaction-

ln(\frac{k_{2}}{k_{1}})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})

where k_{2} and k_{1} are rate constants of reaction at T_{2} and T_{1} temperatures (in kelvin) respectively.

E_{a} is activation energy of reaction.

Here T_{1}= 600 K , k_{1}= 6.1\times 10^{-8}s^{-1}

T_{2}= 725 K, E_{a}= 262 kJ/mol and R = 8.314 J/(mol.K)

So plugin all the values in the above equation-

ln(\frac{k_{2}}{6.1\times 10^{-8}s^{-1}})=\frac{262\times 10^{3}J/mol}{8.314J/(mol.K)}\times (\frac{1}{600K}-\frac{1}{725K})

So, k_{2} = 5.2\times 10^{-4}s^{-1}

Hence rate constant at 725 K is 5.2\times 10^{-4}s^{-1}

7 0
4 years ago
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