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Lubov Fominskaja [6]
3 years ago
9

P.S. This is a science question not a chemistry.

Chemistry
1 answer:
Vesnalui [34]3 years ago
3 0

Answer:

B) When the tilt of the Earth causes the sun’s energy/light to be spread out over a larger area, temperatures are ______warmer______ causing _____Summer_______ to occur.

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In a person's body, sulfur oxides combine with water vapor in the BLANK to form sulfuric acid.
Scilla [17]
This answer would be a blood vessels
7 0
3 years ago
Pressure of the liquid depends upon the height of the liquid column.
Mandarinka [93]

Answer:

The statement is true.

Explanation:

&he pressure exerted by a liquid depends on the height of the liquid column. Where h is height and ρ is density. As the height increases, pressure will also increase.

5 0
3 years ago
A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H 4, 3ClCH
bearhunter [10]

Answer:

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = 1.36 x 10^-3

Explanation:

The reaction

CICH2COOH  ⇄ H+ (aq) + CICH2COO- (aq)

The initial concentration of CICH2COOH is 0.10 M , chloroacetic acid is 11.0% ionized.

let us calculate Ka

First , find change in concentration

since , 11% ionized

change in concentration = 0.10 X 11% = 0.011 M

Initial Concentration of CICH2COOH = 0.10 M

change in concentration of CICH2COOH = - 0.011 M

Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m

Initial Concentration of CICH2COO- = 0 M

change in concentration of CICH2COO- = + 0.011 M

Equilibrium Concentration of CICH2COO- = 0.011 M

Initial Concentration of H+ = 0 M

change in concentration of H+ = + 0.011 M

Equilibrium Concentration of H+ = 0.011 M

Therefore,

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = [H+][CICH2COO-] /[CICH2COOH]

Ka = (0.011 * 0.011) / (0.089)

Ka = 1.36 x 10^-3

6 0
3 years ago
Which of the following is the only metal that is liquid at room temperature?
otez555 [7]
The answer's D. Mercury
3 0
4 years ago
Read 2 more answers
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
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