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3 years ago

If a position time graph has a constant slope, what can you infer about the acceleration?

Chemistry

1 answer:

gogolik [260]3 years ago

5 0

Answer:

Acceleration is zero.

Explanation:

The slope of a position time graph gives the velocity of the body.

If the slope is constant means the velocity is constant.

Now, acceleration is the measure of the change in velocity of a body over a given time interval.

So, the acceleration of a body is directly proportional to the change in velocity of the body.

If there is no change in velocity, this means that the acceleration of the body is zero.

Here, the slope is a constant implying that the velocity is a constant. So, there is no change in velocity. This implies that the acceleration is zero for the body in the given time interval.

Thus, if a position time graph has a constant slope, one can infer that the acceleration is zero.

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In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌

Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of $CO$ = 0.35 mole

Volume of container = 1 L

Initial concentration of $CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M$

Initial moles of $H_2O$ = 0.40 mole

Volume of container = 1 L

Initial concentration of $H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M$

equilibrium concentration of $CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M$ [/tex]

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial conc. 0.35 M 0.40M 0 0

At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}$

$K_c=\frac{x\times x}{(0.40-x)(0.35-x)}$

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

$K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}$

$K_c=0.70$

Thus the value of the equilibrium constant is 0.70.

5 0

3 years ago

a solution must be at a higher temperature than a pure solvent to boil. what colligative property can be employed to achieve thi

Kazeer [188]

Boiling-point elavation.

6 0

3 years ago

Read 2 more answers

Is this correct im so confused.

inysia [295]

Answer:

<em>the last option!!!!</em>

Explanation:

because its right

4 0

3 years ago

At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con

djyliett [7]

0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l

4 0

3 years ago

Magnesium reacts with Oxygen gas forming Magnesium oxide as shown;

LuckyWell [14K]

Molar mass:

O2 = 16 x 2 = 32.0 g/mol Mg = 24 g/mol

<span>2 Mg(s) + O2(g) --->2 MgO(s)
</span>
2 x 24.0 g Mg -------------> 32 g O2
5.00 g Mg -----------------> ( mass of O2)

mass of O2 = 5.00 x 32 / 2 x 24.0

mass of O2 = 160 / 48

= 3.33 g of O2

hope this helps!

5 0

3 years ago