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3 years ago

Why does a warm front usually bring a light and steady rain?

1 answer:

3 0

Warm front Forms when a moist, warm airmass slides up and over a cold air mass. As the warm airmass rises, it condenses into a broad area of clouds. A warmfront brings gentle rain or light snow, followedby warmer, milder weather.

<h2>Mark as brainlist please</h2>

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Your friend claims that weather predictions are just guesses. What would you say to change your friend's mind? Include at least

Nookie1986 [14]

Answer:

I would tell them to mind their business

6 0

3 years ago

Which element requires the least amount of energy to remove the outer-most electron from a gaseous atom in the ground state?

dsp73

<span>the element that requires the least amount of energy to remove the outer-most electron from a gaseous atom in the ground state is s</span><span>odium</span>

8 0

3 years ago

Read 2 more answers

222 g of MTBE (CO(CH3)4) are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). Assume that the de

Eva8 [605]

Answer:

Explanation:

From the information given:

(a)

$Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}$

$Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}$

$Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L$

(b)

$number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}$

(c)

$w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%$

$where; \\ \\ mass \ of \ (RFG) = 2L \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g$

∴

$w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}$

(d)

$Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}$

$Volume \ of \ MTBE = 300 \ mL\\$

∴

$v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}$

(e)

$From \the \ given \ information; \\ \\ 2.5184 \ moles\ of \ MTBE contain \ 2.5184 \ mole of oxygen$

∴

$mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g$

∴

$\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%$

$\% w/w == 2.88\%$

3 0

3 years ago

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Which of these processes indicates that smelting copper is a chemical change

BlackZzzverrR [31]

The correct answer is D:

it involves breaking molecular bonds between copper compounds .

The explanation :

-when we melt a copper it is a physical change because the substance is still copper and have the same shape.

- but for example Burning a copper it is a chemical change. Fire activates a chemical reaction between copper and oxygen.

-The oxygen in the air reacts with the copper and the chemical bonds are broken.

- the chemical change is changing the other compound bonded to the copper atoms.

So, the correct answer is D

5 0

3 years ago

Read 2 more answers

At 3 atm pressure the volume of gas is 1200ml. What pressure is required to reduce the volume of the gas to 300ml if the tempera

Rufina [12.5K]

States that when a gas is held at a constant temperature and mass in a closed container, the volume and pressure vary inversely. The equation to use is P1V1=P2V2.

Given
V1=200mL×1L1000mL=0.2 L
P1=700 mmHg
V2=100mL×1L1000mL=0.1 L

Unknown
P2

Equation
P1V1=P2V2

Solution
Rearrange the equation to isolate P2 and solve.

P2=P1V1V2

P2=(700mmHg×0.2L)0.1L=1400 L, which must be rounded to 1000 L because all of the measurements have only one significant figure.

7 0

3 years ago