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Lesechka [4]
3 years ago
13

Give an example of a time when you would want to increase friction.

Chemistry
1 answer:
Snezhnost [94]3 years ago
6 0

1) you want to increase friction when it gets cold. If you're outside and it's really cold, you're going to rub your hands to warm them up, therefore friction is increasing

I'm not do sure about decreasing.

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A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0
3 years ago
What are the products in a sodium and water reaction?
UkoKoshka [18]

Answer:

Sodium Hydroxide

Explanation:

7 0
3 years ago
Which of the following symbolic representations of matter signifies a mixture of elements?
andreev551 [17]
D only..

A is pure compound
B is mixture of compound and element
C is mixture of compounds
D is mixture of elements
7 0
3 years ago
Please help me with this homework
KengaRu [80]

Answer:

The answer is C density.

8 0
3 years ago
How are half life and radioactive decay related
hoa [83]

Answer : Half life and radioactive decay are inversely proportional to each other.

Explanation :

The mathematic relationship between the half-life and radioactive decay :

N=N_oe^{-\lambda t}              ................(1)

where,

N = number of radioactive atoms at time, t

N_o = number of radioactive atoms at the beginning when time is zero

e = Euler's constant = 2.17828

t = time

\lambda = decay rate

when t=t_{1/2} then the number of radioactive decay become half of the initial decay atom i.e N=\frac{N_o}{2}.

Now substituting these conditions in above equation (1), we get

\frac{N_o}{2}=N_oe^{-\lambda t_{1/2}}

By rearranging the terms, we get

\frac{1}{2}=e^{-\lambda t_{1/2}}

Now taking natural log on both side,

ln(\frac{1}{2})=-\lambda \times t_{1/2}

By rearranging the terms, we get

t_{1/2}=\frac{0.693}{\lambda}

This is the relationship between the half-life and radioactive decay.

Hence, from this we conclude that the Half life and radioactive decay are inversely proportional to each other. That means faster the decay, shorter the half-life.

3 0
3 years ago
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