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nikdorinn [45]
3 years ago
6

In Europe gasoline is often sold in liters. If a typical fill-up requires 11 gal of gas, how many liters are necessary?

Chemistry
1 answer:
egoroff_w [7]3 years ago
3 0

Answer:

41,64 liters

Explanation:

In order to change the units we need to know the equivalence of said units.

1 gal of gas equals to 3,78541 liters.

We can use cross-multiplication in order to convert it into liters:

\frac{1 gal}{3.78541 L} = \frac{11 gal}{X}  \\\\X= \frac{3.78541 L x 11 gal}{1 gal}

A typical fill-up requires 41,64 liters.

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A boy swings a rubber ball attached to a string over his head in a horizontal, circular path. The piece of string is 1.15 m long
gregori [183]

Answer:

v = 16.49 m/s

Explanation:

Given that,

Length of the string, l = 1.15 m

The ball makes 137 complete turns each minute.

We know that, 1 turn = 6.28 rad

137 turns = 860.79 rad

1 min = 60 s

\omega=\dfrac{860.79\ rad}{60\ s}\\\\=14.34\ rad/s

We need to find the tangential velocity of the ball. It can be given by

v=r\omega\\\\=1.15\times 14.34\\\\v=16.49\ m/s

So, the tangential velocity of the ball is 16.49 m/s.

5 0
3 years ago
Kinematics equations are used only by physicists and people who work in
nevsk [136]

Answer:

FALSE

Explanation:

false. kinematics equations are used for many purposes.it is used to drive equations and to find the motion of an object. but it is used more in physics rather than maths.

5 0
3 years ago
Read 2 more answers
A possible mechanism for the gas phase reaction of NO and H2 is as follows: Step 1 2NO N2O2 Step 2 N2O2 + H2 N2O + H2O Step 3 N2
mel-nik [20]

Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: 2NO \Leftrightarrow N_{2}O_{2}    (fast)

Rate expression for step 1 is as follows.

               Rate = k [NO]^{2}

Step 2: N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O    (fast)

Here, N_{2}O_{2} is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

4 0
3 years ago
Describe how the introduction of a new Predator can cause a species
Alexus [3.1K]
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5 0
3 years ago
Create the Equation: What is the Percent Yield of Ammonia (NH3) if 11.8 g is recovered in a reaction with 7.02 x 10^23 molecules
insens350 [35]

Answer:

Explanation:

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a

100

%

yield.

The balanced chemical equation

N

2

(

g

)

+

3

H

2

(

g

)

→

2

NH

3

(

g

)

tells you that every

1

mole of nitrogen gas that takes part in the reaction will consume

3

moles of hydrogen gas and produce

1

mole of ammonia.

In your case, you know that

1

mole of nitrogen gas reacts with

1

mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react

what you need

3 moles H (sub 2)

>

what you have

1 mole H (sub2)

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.

So, the reaction will consume

1

mole of hydrogen gas and produce

1

mole H

2

⋅

2 moles NH

3

3

moles H

2

=

0.667 moles NH

3

at

100

%

yield. This represents the reaction's theoretical yield.

Now, you know that the reaction produced

0.50

moles of ammonia. This represents the reaction's actual yield.

In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every

100

moles of ammonia that could theoretically be produced.

You know that

0.667

moles will produce

0.50

moles, so you can say that

100

moles NH

3

.

in theory

⋅

0.50 moles NH

3

.

actual

0.667

moles NH

3

.

in theory

=

75 moles NH

3

.

actual

Therefore, you can say that the reaction has a percent yield equal to

% yield = 75%

−−−−−−−−−−−−−

or 75 moles NH sub3

I'll leave the answer rounded to two sig figs.

5 0
3 years ago
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