What is the negatively charged sub-atomic particle of an atom?

Can atoms be created or destroyed during a chemical reaction?

ValentinkaMS [17]

Answer:

No.

Explanation:

During chemical reaction, atomes cannot be created or destroyed, they can only react together to form Element or Compound at the Product side.

7 0

2 years ago

A deck of cards has dimensions 8.9cm x 5.72cm x 1.82 cm. What is the volume of the deck in cubic centimeters?

arsen [322]

Answer:

92.65256 cm^3

Explanation:

To find this, we can simply multiply all three dimensions to get the answer in cubic centimeters, and we get the answer above. If you want to be more specific, we can go by the sigfig rule and the answer would be rounded to 93 cm^3.

5 0

2 years ago

rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is

Sunny_sXe [5.5K]

Complete Question

The rate of a certain reaction is given by the following rate law:

$rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$ ?

What is the reaction order in $I_2$ ?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in $H_2$ is n = 1

The reaction order in $I_2$ is m = 1

The overall reaction order z = 2

When the hydrogen is double the the initial rate is $rate_n = 4.0*10^{-4} M/s$

The rate constant is $k = 35.23 \ M^{-1} s^{-1}$

Explanation:

From the question we are told that

The rate law is $rate = k [H_2][I_2]$

The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be n and for $I_2$ be m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is n=m=1

The overall reaction order is

$z = n +m$

$z =1 +1$

$z =2$

At $rate = 2.0 *10^{4} M /s$

$2.0*10^{4} = k [H_2] [I_2] ---(1)$

= > $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

$rate = k [2H_2] [I_2] ----(2)$

=> $k = \frac{rate_n }{ [2H_2] [I_2]}$

So equating the two k

$\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

=> $rate_n = 4.0*10^{-4} M/s$

So when

$rate_x = 52.0 M/s$

$[H_2] = 1.8 M$

$[I_2] = 0.82 \ M$

We have

$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

$k = 35.23 M^{-2} s^{-1}$

3 0

2 years ago

What is the ∆G for the following reaction between O3 and O2? 3O2(g) 2O3(g) Given: O3: ∆H = 285.4 kJ ∆S = -137.14 J/K)

Natali [406]

We’ll be using the equation:
dG = dH - TdS (replace ‘d’ with triangle)

I’m going to assume 0 degrees Celsius.
At 0 C (273 K):

dG = dH - TdS
dG = (285,400 J) - (273 K)(-137.14 J/K)
dG = 285,400 J + 37,439.2 J
dG = 322,839.2 J or 322.84 kJ

The dG of this reaction is +322.84 kJ. This reaction is not considered spontaneous.

This answer, in this instance, would be D. If the temperature used in the question is not 0 degrees C, replace the temperature that I used for calculation with the Kelvin temperature given in the problem (K = C + 273), and simplify to find the answer.

6 0

2 years ago

Define self reduction reaction.

alexira [117]

Answer:

Self-reduction is a reaction in which no external reducing agent is required. It is called self-reducing because copper oxide itself reduced the copper sulfide to give copper and sulfur dioxide.

7 0

2 years ago