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3 years ago

How many cm are in 7 km?

Chemistry

2 answers:

kipiarov [429]3 years ago

7 0

Answer:

700000cm

Explanation:

1 km consists of 1000m and 1 m consists of 100 cm so 7×1000×100=700000cm

nikklg [1K]3 years ago

5 0

Answer:

70000 cm in 7km

Explanation:

There are 1000 m and 100 cm in 1 m in 1 km, so to find how many cm are in 7 km you need to do 10000 x 7. You get 70000

Have a nice day!

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If a 58 g sample of metal at 100 c is placed into calorimeter containing 60g of water at 18 c , the temperature of the water inc

tatiyna

Answer:

The water will absorb 1004.16 Joule of heat

Explanation:

Step 1: Data given

Mass of the metal = 58.00 grams

Temperature of the metal = 100.00 °C

Mass of water = 60.00 grams

Temperature of water = 18.00 °C

Final temperature = 22.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the amount of heat absorbed by the water in joules

Q = mass *specific heat *ΔT

⇒ with Q = the heat absorbed by water

⇒ with mass of water = 60.00 grams

⇒ with specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature of water = T2 - T1 = 22 - 18 = 4.0 °C

Q = 60.00 * 4.184 J/g°C * 4.0 °C

Q = 1004.16 J

The water will absorb 1004.16 Joule of heat

6 0

3 years ago

If the volume occupied by 0.500 mol of nitrogen gas at 0°C is 11.2 L, then the volume occupied by 2.00 mol of nitrogen gas at th

Paraphin [41]

Avagadros law states that volume of gas is directly proportional to number of moles of gas at constant pressure and temperature.
$\frac{V1}{n1} = \frac{V2}{n2}$
where V -volume , n - number of moles
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substituting these values in the equation
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answer is C. 44.8 L

5 0

3 years ago

Can you add more points to question pls

Zarrin [17]

what do you mean? explain what you r trying to dsay

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3 years ago

For the balanced equation shown below, if the reaction of 0.112 grams of

Alekssandra [29.7K]

Answer:

The answer is 74.5%.

Explanation:

As we know that % yield= $\frac{actual yield}{theoretical yield}$ x 100%.

Therefore,

Step 1 Calculate Theoretical yield:

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