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elena-s [515]
3 years ago
15

If the relative humidity is high and the temperature drops, water vapor may

Chemistry
1 answer:
meriva3 years ago
6 0

If the water vapor content stays the same and the temperature drops, the relative humidity increases. If the water vapor content stays the same and the temperature rises, the relative humidity decreases. This is because colder air doesn't require as much moisture to become saturated as warmer air.

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If a black ball is denser than a white ball of the same size, what does the black ball have??
yKpoI14uk [10]

Answer: The black ball will have more mass than the white ball.

Explanation: Density of a substance is defined as the ratio of mass and volume occupied by the substance.

Mathematically,

Density=\frac{Mass}{Volume}

We are given that the two balls are of same size, which means that the volume of both the balls are same. We are also given that the black ball is more dense than the white ball, which means that the black ball will have more mass.

As, Density\propto Mass

7 0
3 years ago
BALANCE THE EQUATION:<br> C4H10 + O2 = CO2 + H2O
Tasya [4]

2 C 4 H 10 ( g ) +  13 O 2 ( g )  =  8 C O 2 ( g ) +  10 H 2 O ( g )

6 0
3 years ago
9. What happens to the thermal energy that is added to the water between 4 and 11 minutes?
Sedaia [141]

Answer:

C. It causes the particles to move faster

8 0
3 years ago
Why does a solid have a definite shape and volume?
erica [24]

Answer:

B

Explanation:

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8 0
2 years ago
For a first-order reaction, A → B, the rate coefficient was found to be 3.4 × 10-4 s-1 at 23 °C. After 5.0 h, the concentration
Illusion [34]

Answer:

the original concentration of A = 0.0817092  M

Explanation:

A reaction is considered to be of first order it it strictly obeys the graphical equation method.

k_1 = \dfrac{2.303}{t}log \dfrac{a}{a-x}

where;

k = the specific rate coefficient  = 3.4 × 10⁻⁴ s⁻¹

t = time   = 5.0 h = 5.0 × 3600 = 18000 seconds

a = initial concentration = ???

a - x = remaining concentration of initial concentration at time t = 0.00018 mol L⁻¹

3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}

3.4 \times 10^{-4}= 1.27944 \times 10^{-4} \times log \dfrac{a}{0.00018}

\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}=   log \dfrac{a}{0.00018}

2.657=   log \dfrac{a}{0.00018}

10^{2.657}= \dfrac{a}{0.00018}

453.94 = \dfrac{a}{0.00018}

a =453.94 \times 0.00018

a = 0.0817092  M

Thus , the original concentration of A = 0.0817092  M

8 0
3 years ago
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