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3 years ago

When warm air is forced upward along a cold front what might happen?

1 answer:

mote1985 [20]3 years ago

8 0

Answer: With the cold front, warm air is rapidly forced upward (like the shavings) in advance of the actual front (the “cutter”), creating towering cumulus clouds, some hard showers and quite possibly a few gusty thunderstorms followed by a push of cooler and drier air in its wake.

Explanation:

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An air sample contains 0.038% CO2. If the total pressure is 758 mmHg, what is the partial pressure of CO2?

Helga [31]

<span>(0.038 / 100) * 763 = 0.29 mmHg
hope it helps
</span>

7 0

3 years ago

Can anyone give the ans<br><br><br><br><br><br><br><br>please<br><br><br><br><br><br>...

Burka [1]

Answer:

..................mm.m

8 0

4 years ago

How much rust is produced with 1.5 kg of Fe reacts with water

ZanzabumX [31]

Answer:

2071g or 2.071kg of rust (Fe3O4)

Explanation:

Step 1:

The balanced equation for the reaction.

3Fe + 4H2O —> Fe3O4 + 4H2

Step 2:

Determination of the mass of Fe that reacted and the mass of the Fe3O4 produced from the balanced equation.

Molar Mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 3 x 56 = 168g

Molar mass of Fe3O4 = (56x3) + (16x4) = 232g/mol

Mass of Fe3O4 from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

168g of Fe reacted and 232g of Fe3O4 was produced.

Step 3:

Determination of the mass of rust (Fe3O4) produced when 1.5kg ( i.e 1500g) of Fe reacted.

This is illustrated below:

From the balanced equation above,

168g of Fe reacted to produce 232g of Fe3O4.

Therefore, 1500g of Fe will react to produce = (1500x232)/168 = 2071g of Fe3O4.

From the calculations made above, 2071g or 2.071kg of rust (Fe3O4) is produced.

6 0

3 years ago

How many acetone molecules are in a bottle of acetone with a volume of 445 mL ? (density of acetone = 0.788 g/cm3).

Mkey [24]

<h2>Hello!</h2>

The answer is: $3.63x10^{24}particles$

First, we need to calculate the molecular weight of the acetone (CH3COCH3)

So,

$C=12.011g/mol\\O=15.99g/mol\\H=1.008g/mol$

Then,

$CH3COCH3=12.011+(1.008)*3+12.011+15.999+12.011+(1.008)*3=58.08g/mol$

Second, we need to calculate the mass of the acetone in the bottle

We need to remember that $1cm^{3}=1mL$

So,

$445cm^{3}*\frac{0.788g}{cm^{3}}=350.66g$

Third, we need to calculate the number of moles:

$350.66g*\frac{1mol}{58.08g}=6.03moles$

Fourth, we need to calculate the number of atoms:

Remember, $1 mol=6.022x10^{23}particles$

Therefore,

$6.03moles*\frac{6.022x10^{23}particles }{1mol}=3.63x10^{24}particles$

Have a nice day!

4 0

4 years ago

Read 2 more answers

How much energy is lost to condense 300. grams of steam at 100.C?

True [87]

Energy lost to condense = 803.4 kJ

<h3>Further explanation</h3>

Condensation of steam through 2 stages:

1. phase change(steam to water)

2. cool down(100 to 0 C)

1. phase change(condensation)

Lv==latent heat of vaporization for water=2260 J/g

$\tt Q=300\times 2260=678000~J$

2. cool down

c=specific heat for water=4.18 J/g C

$\tt Q=300\times 4.18\times (100-0)=125400$

Total heat =

$\tt 678000+125400=803400~J$

3 0

3 years ago