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2 years ago

11

Write a balanced chemical equation for this reaction by completing the reaction below. Add the missing products and coefficients

. Do not include phases.

1 answer:

Pie2 years ago

6 0

To balance a chemical equation you must pay attention that the algebraic equation has the same number of element atoms in reactant and the product.

A chemical equation is a term to refer to the symbolic description of a chemical reaction, that is, the written representation of symbols. For example:

2H₂ + O₂ ---> 2H₂O

The balance of an equation can be related to the law of conservation of matter, that is, the number of atoms of each element in the reactants is in the result.

Note: This question is incomplete, because the information some information is missing.

Learn more in: brainly.com/question/12271256

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6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets

Nata [24]

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: $V_{1}$ = 0.20 mL, $T_{1} = 27^{o}C$

$V_{2}$ = ?, $T_{2} = 177^{o}C$

Formula used to calculate new volume is as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL$

Thus, we can conclude that the new volume of cake is 1.31 mL.

5 0

3 years ago

A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L

NeTakaya

Answer:

$\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}$

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

$P_1V_1=P_2V_2$

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

$25.0 \ L * 2.05 \ atm = P_2V_2$

The volume is decreased to 14.5 liters, but the pressure is unknown.

$25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L$

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}$

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2$

The units of liters cancel.

$\frac {25.0 * 2.05 \ atm }{14.5 }=P_2$

$\frac {50.25\ atm }{14.5 }=P_2$

$3.53448276 \ atm = P_2$

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

$3.53 \ atm \approx P_2$

The new pressure is approximately <u>3.53 atmospheres.</u>

8 0

3 years ago

Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.

Ket [755]

Answer:

c. 2,2-dichloropentane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:

$CH\equiv C-CH_2-CH_2-CH_2$

Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:

$CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2$

Whose name is 2,2-dichloropentane.

Regards!

8 0

3 years ago

What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?

xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

$K_{a}=1.8\times 10^{-5}$

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

$\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}$

The expression for dissociation constant of acid is:

$K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}$

$1.8\times 10^{-5}=\frac{x^2}{1.8-x}$

$1.8\left(1.8-x\right)=100000x^2$

Solving for x, we get:

<u>x = 0.00568 M</u>

Percentage ionization = $\frac{0.00568}{1.8}\times 100=0.32 \%$

<u>Option B is correct.</u>

8 0

3 years ago

2. Which of the following hydrocarbons would you expect to have the

raketka [301]

Answer:

C

6

H

14

Explanation:

5 0

2 years ago