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Goshia [24]
3 years ago
11

When driving a truck,is fuel crucial?

Chemistry
1 answer:
Natali [406]3 years ago
6 0

Answer:

The answer is yes.

Explanation:

When driving any type of motorized vehicle, fuel is importatn. Fuel is what it runs on, so without it it wouldn't run. With trucks a lot more is needed because trucks are bigger, and tend to carry more than a little car does.

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Select all that apply. Using your periodic table of the elements which chemical symbols indicate an element?
tatuchka [14]

Answer:

O, I, Ca

Explanation:

4 0
3 years ago
Gold is currently trading at very high price. Suppose that gold is selling for around $1860/ounce. How
Ugo [173]

Answer:

The answer is "3.81041978"

Explanation:

\to 1 \ OZ= 28.349523125 \ grams\\\\

              =28.349523125\times {1000} \ miligrams\\\\= 28349. 5231  \ miligrams\\

In 1860 = 28349.5231 \ miligrams\\

\to In \$ \ 1 = \frac{28349.5231}{1860} \ \ miligrams\\

             = 15.2416791 \ miligrams

\ In \  1 \ quarter =  \$ \ 0.25

\to \$ \ 0.25 =  15.2416791  \times 0.25 \  miligrams\\

              = 3.81041978

3 0
3 years ago
The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample
IgorLugansk [536]

Answer:

6.2 calories

Explanation:

Data Given:

change in temperature = 20 °C

specific heat of gold = 0.031 calories/gram °C

mass of gold = 10.0 grams

Amount of Heat = ?

Solution:

Formula used

             Q = Cs.m.ΔT

Where:

Q = amount of heat

Cs = specific heat of gold = 0.031 calories/gram °C

m = mass

ΔT = Change in temperature

Put values in above equation

                Q = 0.031 calories/gram °C x 10.0 g x 20 °C

                Q = 6.2 calories

So option A is correct = 6.2 calories

6 0
3 years ago
If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it
Nitella [24]

The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

NO_2\rightarrow NO+\frac{1}{2}O_2

The reaction is second order for NO_2 with a rate constant of 0.543M^{-1}s^{-1} at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M

a) 1.01    b) 5.19     c) 0.299      d) 0.0880     e) 3.34

<u>Answer:</u> The time taken is 5.19 seconds

<u>Explanation:</u>

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.543M^{-1}s^{-1}

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.150 M

[A]_o = Initial concentration = 0.260 M

Putting values in above equation, we get:

0.543=\frac{1}{t}\left (\frac{1}{(0.150)}-\frac{1}{(0.260)}\right)\\\\t=5.19s

Hence, the time taken is 5.19 seconds

6 0
3 years ago
Why are atoms of carbon-14(C-14) unstable?
vfiekz [6]
They are a radioactive form of carbon, they are bad bc they have too many neutrons for its six protons, making it unstable
3 0
3 years ago
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