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tamaranim1 [39]

2 years ago

8

iron combines with 4.00 g of copper (11) nitrate to form 6.01 g of Iron (l) nitrate and 0.400 g copper metal. how much iron did

it take to convert to Cu(NO3)2?

1 answer:

kvasek [131]2 years ago

7 0

Answer:

idkkk

Explanation:

haaaaa

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

2 years ago

Categorize the following materials: bronze alloy, mouse growing an ear on its back, porcelain dentures

masha68 [24]

Bronze alloy and porcelain dentures

5 0

2 years ago

What features of this model will help Armando answer the question?

Scrat [10]

Answer:

The adjustable legs and the table of sand.

<em>Note:The question is incomplete. The complete question is given below.</em>

Using Models to Answer Questions About Systems

Armando’s class was looking at images of rivers formed by flowing water. Most of the rivers were wide and shallow, but one river was narrow and deep. Armando’s class thinks that this river is narrow and deep because:

the hill that the water flowed down was very steep, or
the sand grains that the water flowed through were very small.

Armando designed the model below to try to answer the question: Why is this river so narrow and deep?

Explanation:

The model designed by Armando will be helpful to answer the question because of the following features it possesses:

1. An adjustable leg- since one of the hypotheses put forward by the class to explain why the river was narrow and deep was that the hill that the water flowed down was very steep, the adjustable legs can be lowered or raised in order to make the slope shallower or steeper so that their hypothesis can be tested.

2. A table of sand- the table of sand serves as the streambed. By adjusting the size of the sand grains to be larger or smaller, the students will be able to to test their second hypothesis that the small size sand grains that the water flowed through was the reason for the river to be narrow and deep.

The results of their experiments will enable them to come to a conclusion.

5 0

2 years ago

Read 2 more answers

Suppose you have a system made up of water only, with the container and everything beyond being the surroundings. Consider a pro

Airida [17]

Answer:

Yes

Explanation:

The possibility of evaporating and condensing is a proof of reversible reaction

8 0

2 years ago

Read 2 more answers

Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

7 0

2 years ago