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3 years ago

Take car traveling with constant speed travels at 150 km in 7200 a. what is the speed of the car?

1 answer:

5 0

Find the speed of the car in which it travels at 150 km in 7200 seconds.
This is an easy question, you just need to follow the given formula as always.
Since the time and the distance is already given, we are now looking for the speed
=> time = 7200 seconds
=> distance = 150km
d = speed x time
S = distance / time
s = 150 km / 7200 seconds
s = 0.021 km / seconds.
Thus, the car travels for about 0.021 km per seconds.

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How can you tell the number of covalent bonds an element can form?

Sergio039 [100]

The number of bonds for a neutral atom is equal to the number of electrons in the full valence shell (2 or 8 electrons) minus the number of valence electrons. This method works because each covalent bond that an atom forms adds another electron to an atoms valence shell without changing its charge.

3 0

3 years ago

Reaction is

-Dominant- [34]

According to the task, you are proveded with patial pressure of CO2 and graphite, and here is complete solution for the task :
At first you have to find n1 =moles of CO2 and n2 which are moles of C
The you go :
 $CO2(g) + C(s) \ \textless \ =\ \textgreater \ 2CO(g)$

n1 n2 0
-x -x +2x
$n1-x n2-x 2x Kp=P^2 (CO)/Pco2$
After that you have to use the formula $PV=nRT$
$Kp=(2x)^2/n1-x$
Then you have to solve x, and for that you have to use RT/V
And to find total values: $P(total)=n(total)*R*T/V$
I am absolutely sure that this would be helpful for you.

7 0

3 years ago

1. A sample of oxygen is collected over water at 22 ° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor

Rom4ik [11]

Answer: The partial pressure of the dry oxygen is 742 torr

Explanation:

Dalton's Law of Partial Pressure states that the total pressure exerted by a mixture of gases is the sum of partial pressure of each individual gas present. Thus $P(total)=P_1+P_2 .........$

Given; Total pressure = 762 torr

partial pressure of water = 19.8 torr

partial pressure of dry oxygen = ? torr

Total pressure = partial pressure of water + partial pressure of dry oxygen

762 torr = 19.8 torr = partial pressure of dry oxygen

partial pressure of dry oxygen = 742 torr

The partial pressure of the dry oxygen is 742 torr

8 0

3 years ago

Will mark brainliest :) how many moles of a solute is present in 4.00 L of an 8.30 M solution?

andreev551 [17]

The answer is 33.2 moles of a solute is present in 4.00 L of an 8.30 M solution , Option A is correct .

<h3>What is Molarity ?</h3>

Molarity is defined as the amount of solute (in moles)in per litre of solution.

It is also known as molar concentration of a solution , It is expressed in mol/l

$\rm M =\dfrac{moles \;of\; solute}{Volume \;of \;solution} \\\\\\\rm M = \dfrac{n}{V} \\$

We can rearrange this equation to get the number of moles:

n= M * V

The molarity of solution is 8.3 M and the volume given is 4 litres

the moles will be n = 8.30 * 4 = 33.2 moles

Therefore 33.2 moles of a solute is present in 4.00 L of an 8.30 M solution , Option A is correct .

To know more about molarity

brainly.com/question/2817451

#SPJ1

5 0

2 years ago

The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63x10^-3 at 527 C. Calculate the equilibrium partial

Studentka2010 [4]

Answer: The partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

Explanation:

We are given:

$K_c=4.63\times 10^{-3}$

$p_{COCl_2}=0.760atm$

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $4.63\times 10^{-3}$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $527^oC=527+273=800K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304$

The chemical reaction for the decomposition of phosgene follows the equation:

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

At t = 0 0.760 0 0

At $t=t_{eq}$ 0.760-x x x

The expression for $K_p$ for the given reaction follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

We are given:

$K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x$

Putting values in above equation, we get:

$0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656$

Negative value of 'x' is neglected because partial pressure cannot be negative.

So, the partial pressure for the components at equilibrium are:

$p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm$

Hence, the partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

6 0

3 years ago