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2 years ago

The chart below shows examples of solids, liquids, gases, and plasma.

2 answers:

4 0

The word that is in an incorrect place in the chart is HAIL.
The state of matter of hail is not plasma; hail is a solid at room temperature. Plasma refers to the state of matter in which an ionized gas have approximately equal number of positively and negatively charged ions. Thus, plasma is an ionized gas. Plasma is considered to be the fourth state of matter. <span />

ElenaW [278]2 years ago

3 0

Its hail because hail is a solid.

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Determine the number of moles of carbon dioxide gas, water, and sodium chloride formed by the reaction of 42.0 grams of sodium b

brilliants [131]

Answer: .45 moles of CO2, H20,and NaCl

Explanation:

5 0

3 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

2 years ago

A student wants to remove the salt from a mixture of sand and salt in order to get only pure sand. He adds water to the mixture.

raketka [301]

Answer:

Here is one way: Add water to the mixture. Only the sugar dissolves. This is a physical change.

Explanation:

The sugar would dissolve in water. You could then pour off the solution and wash the remaining sand with a bit more water. Heat the water to evaporate it from the sugar, and the two are separated.

5 0

2 years ago

Read 2 more answers

What is the standard enthalphy change ΔHo, for the reaction represented above? (ΔHof of C2H2(g) is 230 kJ mol-1; (ΔHof of C6H6(g

wolverine [178]

Answer:

-608KJ/mol

Explanation:

3 C2H2(g) -> C6H6(g)

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= ΔHC6H6 - 3ΔHC2H2

ΔHrxn = 83 - 3(230)