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3 years ago

Which of the following is a true

Chemistry

1 answer:

Katarina [22]3 years ago

5 0

❤️Hello!❤️ The answer is A. Energy can change from one form to another. Hope this helps! ↪️ Autumn ↩️

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For the balanced equation shown below, what would be the limiting reagent if 93.9grams

svlad2 [7]

Answer: A. C2H3F

Explanation:

5 0

2 years ago

The products

alexandr1967 [171]

Answer:

The products of self-ionization of water are OH⁻ and H⁺.

Explanation:

The water is self ionized according to the equation:

The ionic product for water (Kw) = [OH⁻][H⁺] = 10⁻¹⁴.

Kw is also called "self-ionization constant" or "auto-ionization constant".

6 0

3 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

3 years ago

Sodium hydroxide, NaOH, reacts with sulfuric acid, H,SO,, in a neutralization reaction to

RSB [31]

Answer:

10.6 grams is approximately 0.10 moles. So you would need about 0.10 moles of sulfuric acid. That converts to about 9.80 grams.

Explanation:

hi girl i also wrote this in my test today so maube i hope it is correct.

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7 0

2 years ago