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3 years ago

How many atoms of phosphorus are in 4.90 mol of copper(II) phosphate?

Chemistry

1 answer:

Nutka1998 [239]3 years ago

8 0

There are 6.02214*10^23 molecules , I'm not so sure about it though.

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Mendeleev's periodic table of elements was created in 1901. O A. True O B. False

LiRa [457]

B. False
It was created in 1869

3 0

3 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

How can we specify / identify a certain electron? (Quantum forces)

Serga [27]

Answer:

electrons are less than 1 amu, they are also negative charged. they are very small too

3 0

3 years ago

Which of the following was not a part of dalton's atomic theory?

zhuklara [117]

Letter D. ☺☺☺☺☺☺☺☺☺☺☺☺☺☺☺

3 0

2 years ago

The value of ΔH for the reaction below is -72 kJ. __________ kJ of heat are released when 1.0 mol of HBr is formed in this react

eimsori [14]

Answer:

36 KJ of heat are released when 1.0 mole of HBr is formed.

Explanation:

By Hess law,

The heat of any reaction ΔH for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction:

H 2 (g) + Br 2 (g) → 2HBr (g) ΔH = -72 KJ

This is the energy released when 2 moles of HBr is formed from one mole each of H2 and Br2.

Therefore, Heat released for the formation of 1 mol HBr would be half of this.

Hence,

ΔHreq = -36 kJ

36 KJ of heat are released when 1.0 mole of HBr is formed.

4 0

3 years ago