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Elena-2011 [213]
2 years ago
14

A Cylinder has a diameter of 20 m and a height of 100 cm.

Chemistry
2 answers:
Alexxandr [17]2 years ago
4 0

Answer:

d

Explanation:

cylinder has a diameter of 20 m and height of 100 cm is 2000mcube

Alenkasestr [34]2 years ago
4 0

Answer:  314 m^3

Explanation:  V = πr²H

H = 1 meter (100cm = 1 meter)

R = 10 meters

V = (3.14)(10m)²(1m)

V = 314 m³

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Amanda [17]
<span>the table say that at 20 degree celcius 88.0g of NANO3 will remain dissolved in 100 gm of H2O so at 20 degree celcius 80.0g of H20 will dissolve (88.0g)x(80g/100g)=70.4g of NANO3 so at 20 degree celcius 86.3g-70.4g= 15.9 gram of NANO3 will come out of solution.</span>
8 0
3 years ago
If the map shows the average high temperature in July for two cities in Texas. like Del Rio=36°c
Sedbober [7]

Answer:

D

Explanation:

Ocean breezes keep coastal galveston cooler than Del Rio, which is inland exposed to southerly winds.

4 0
3 years ago
Which substance/s have covalent bond? Explain your answer.​
Viefleur [7K]
Covalent bonding occurs when electrons are shared between atoms. Double and triple covalent bonds occur when four or six electrons are shared between two atoms, and they are indicated in Lewis structures by drawing two or three lines connecting one atom to another.
4 0
3 years ago
For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁   -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0
3 years ago
What is the name of MnO? Explain how you determined the bond type and the steps you used to determine the naming convention for
xeze [42]

Answer:

Magnesiumoxide

Explanation:

I don't know

5 0
2 years ago
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