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fiasKO [112]
4 years ago
15

Which of the following are steps for balancing chemical equations?

Physics
1 answer:
AveGali [126]4 years ago
5 0

Answer:

C. Recheck the numbers of each atom on each side of the equation

to make sure the sides are equal.

D. Choose coefficients that will balance the equation

Explanation:

In balancing of chemical equation, the number of atoms on both sides must be equal in adherence to the law of conservation of mass.

Using the method of inspection, the equation is first observed to know the relationship between the combining atoms and the resulting ones.

After observing the reaction, put a coefficient that will balance the equation. Then recheck the number of each atom on both side of the equation. One can repeat the process till the equation is balanced.

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It is 5.00 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0 k
8_murik_8 [283]

Answer:

a. Walking burns up more energy.

b. 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

Explanation:

a. We know energy W = Pt where P = power and t = time.

Now for walking, t = d/v where d = distance = 5.00 km and v = speed = 3.00 km/hr and P = 290 W

So, t = d/v = 5.00 km/3.00 km/hr = 5/3 hr = 5/3 × 3600 s = 6000 s

W = Pt = 290 W × 6000 s = 1740000 = 1740 kJ

Now for running, t = d/v where d = distance = 5.00 km and v = speed = 10.00 km/hr

So, t = d/v = 5.00 km/10.00 km/hr = 0.5 hr = 0.5 × 3600 s = 1800 s and P = 700 W

W = Pt = 700 W × 1800 s = 1260000 = 1260 kJ

Since walking burns up 1740 kJ and running burns up 1260 kJ, walking burns up more energy.

b. It burns up 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

4 0
4 years ago
What two major uses does this (H-R)diagram have for astronomers?
EastWind [94]
66666666666666666666666666666666666666666666666666666666666666666666666666666666666666

7 0
3 years ago
It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i
timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is 4.8 \cdot 10^{-6}\,mg

Explanation:

The ration between Uranium mass and total sample mass is: \frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}

For a sample of mass 1.2 mg, the amount of uranium is:

1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg

7 0
3 years ago
A 3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?​
anyanavicka [17]

Answer:

I think the answer is

Explanation:

140N/m/20N =

7m is the answer

8 0
2 years ago
Read 2 more answers
Please help me out i'm so depressed and such a failure
aksik [14]

Answer:

I reckon towards b. Let me know if im right

4 0
3 years ago
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