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drek231 [11]
2 years ago
5

Have you ever tried to mix sugar with a cold drink like tea or lemonade? What happened? Explain why.

Chemistry
1 answer:
Vinvika [58]2 years ago
5 0

Answer:

yes

Explanation:

it wont dissolve because when water is cold its molecules dont separate

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konstantin123 [22]
Magnesium nitride weighs 100.95 g/mol

133/100.95 = 1.32 mols
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2 years ago
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The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its
Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

t_{\frac{1}{2}} =\frac{ln 2}{k}

t_{\frac{1}{2}} = \frac{0.693}{k}

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}

k = \frac{0.693}{5230 years}

k = 1.325 * 10^{-4} yr^{-1}

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

[A] = [A_{0}]e^{-kt}

\frac{[A]}{[A_{0}]} = e^{-kt}

\frac{35}{100} = e^{-(1.325 *10^{-4})t}

ln(0.35) = -(1.325 *10^{-4})(t)

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0
3 years ago
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JulsSmile [24]

Answer:

Either A or D

But i think D

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7 0
3 years ago
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goldfiish [28.3K]
Answer : A and D shows PROCESS of sweating
5 0
2 years ago
An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr
Lera25 [3.4K]

Answer:

\sigma_y\ =210.2\ MPa  

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}

\sigma_y\ =Yield\ strength

d = diameter

K =Constant

\sigma_o\ =material\ constant

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

163=\sigma_o+\dfrac{K}{\sqrt 35}      ------------1

d= 17 μm ,yield strength = 192 MPa

192=\sigma_o+\dfrac{K}{\sqrt 17}           ------------2

From equation 1 and 2

192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}

K=394.53

By putting the values of K in equation 1

163=\sigma_o+\dfrac{394.53}{\sqrt 35}

\sigma_o\ =96.31\ MPa

\sigma_y=96.31+\dfrac{394.53}{\sqrt d}

Now when d= 12 μm

\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}

\sigma_y\ =210.2\ MPa

4 0
3 years ago
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