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Natasha_Volkova [10]
2 years ago
3

A car starts from rest and accelerates along a straight line path in one minute. It finally attains a velocity of 40 meters/seco

nd. What is the car's average acceleration?
A.) 1.5 meters/second2
B.) 1.5 meters/second
C.) 0.66 meters/second2
D.) 0.66 meters/second
Chemistry
1 answer:
Scilla [17]2 years ago
5 0
To answer this question, we will use the equation of motion which states that:
V = u + at where:
V is the final velocity = 40 meters/second
u is the initially velocity. Since the car starts from rest, this means that u = 0
a is the acceleration we need to calculate
t is the time = 1 minute = 60 seconds

Substitute with the givens in the above mentioned equation to get the acceleration as follows:
V = u+at
40 = 0 + 60a
a = 40/60 = 2/3 = 0.667 meter / sec^2

Based on the above calculations, the right choice is:
c. 0.66 meters / second^2

Note that choice D has the right value but the wrong units. Therefore, D is not the correct choice.


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When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

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

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0
2 years ago
After passing through pyruvate dehydrogenase and the citric acid cycle, one mole of pyruvate will result in the formation of ___
mamaluj [8]

Answer:

The answer to be filled in the respective blanks in question is

3 and 1

Explanation:

So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.

Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1

 

7 0
2 years ago
Mr. Rutherford's chemistry class was collecting data in a neutralization study. Each group had 24 test tubes to check each day f
ycow [4]
I’m pretty sure it is A at least that’s what we did at our school to test this
6 0
2 years ago
A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m
ioda
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = \frac{PV}{nT} = constant
\frac{P1 V1}{n T} = \frac{P2 V2}{n T}
==> P1 * V1 = P2 * V2
V2 = \frac{P1 V1}{P2} = \frac{445 0.016}{1225}
V2 = 0.00581 m3 = 5.81 L


7 0
2 years ago
If a chemist analyzes a 3.84g sample containing sand and table sugar, and recovers 1.43g of      sand, what  percent by mass of
disa [49]
3.84 - 1.43 = 2.41
2.41g of table sugar

% mass = ( (mass of element) / (total mass) ) * 100
% mass = (2.41 / 3.84) * 100
% mass = (0.6276) * 100
% mass = 62.76

62.76%
8 0
2 years ago
Read 2 more answers
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