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Greeley [361]
3 years ago
6

Allen is carrying out reactions in a lab. One of the products in all reactions is found to be either a precipitate, a gas, or wa

ter. What is common between all these reactions?
Chemistry
2 answers:
juin [17]3 years ago
4 0

Answer:

Decomposition reaction.

Explanation:

There is formation of a precipitate or evolution of gas or formation of water generally during decomposition reactions.

They involve breaking of a compound into two or more new compounds or one or more element.

Example:

Decomposition of calcium carbonate.

CaCO_{3}--->CaO(s)+CO_{2}(g)

H_{2}SO_{3}--->H_{2}O+SO_{2}(g)

Artist 52 [7]3 years ago
3 0
All decomposition reactions. 
Hope this helps!(: 
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Density measurements were conducted on a 22.5oC sample of water which had a theoretical density of 0.997655 g/ml. A volume of 10
Sophie [7]

Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).

  • (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:

p_{A} =9.98g/10.00mL=0.998g/mL\\p_{B} =15.61g/15.00mL=1.041g/mL\\p_{C} =25.65g/25.00mL=1.026g/mL

  • (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):

p_{average}=\frac{p_{1}+p_{2}+p_{3}}{3}   \\p_{average}=\frac{(0.998+1.041+1.026)g/mL}{3}\\p_{average}=1.022g/mL

  • (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:

%error=\frac{|p_{average}-p_{theoretical}|}{p_{theoretical}} *100

%error=\frac{|1.022g/mL-0.997655g/mL|}{0.997655g/mL}*100

%error=2.44 %

7 0
3 years ago
If you have a 4.6 L of gas in a piston at a pressure of 1.2 atm and compress 2p the gas until its volume is 2.6 L, what will the
Nesterboy [21]

Answer:

2.12atm

Explanation:

Boyle's Law: P1V1 = P2V2

Manipulate to solve for unknown: P2 = P1V1/V2

Substitute values:  P2=(1.2atm)(4.6L)/2.6L

P2 = 2.1230769atm

Round to 3 sig figs to get 2.12atm

3 0
3 years ago
At this lower concentration, about how many extra hydrogen bonds would be needed to hold a and b together tightly enough to form
Scrat [10]

2..............................................

8 0
3 years ago
Read temprature on thermometer i am confusion
balu736 [363]

Answer:

a)22.2°C after adding magnesium

b)17.3°C before adding magnesium

c) 4.9 is change

8 0
3 years ago
(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

6 0
3 years ago
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