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3 years ago

Which weather event usually occurs when a cold front moves into an area?

1 answer:

8 0

I think the answer would be
A) because As a cold front moves into an area, the heavier cool air pushes under the lighter warm air, causing it to rise up into the troposphere. Lifted warm air ahead of the front produces cumulus or cumulonimbus clouds and thunderstorms.

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The atoms of Group 7A elements gain electrons when they form ions. True False

artcher [175]

Answer: True.

Explanation.

The group 7A is actually named group 17.

That group is the halogens: F, Cl, Br, I, At, and Ts (Ts is one of the last elements discovered).

Those elements have 7 valence electrons (notice that it is the same number as the second digit in 17).

The atoms with 7 valence electrons will "easily" gain one electron to get the configuration of the next noble gas (8 valence electrons). That is why these elements gain electrons to form ions.

When atoms gain electrons form anions (negative ions). For example: F(-), Cl(-), Br(-), I(-).

6 0

3 years ago

Read 2 more answers

How many atoms are in 35.12 grams of gold? Please show your work

taurus [48]

Answer: 1.08 x 10^23 atoms

Explanation:

To find the no of moles in 35.12g of Au

no. of moles = mass/atomic mass

= 35.12/197

= 0.18 moles

1 mole of Au = 6.02 x 10^23

0.18 moles of Au = 0.18 x 6.02 x 10^23

=1.08 x 10^23 atoms

3 0

3 years ago

Please help , science tho. by the way i have to do 5

kap26 [50]

Answer:

850

Explanation:

It's between 840 and 860

8 0

2 years ago

The least reactive with oxygen in the list is:. a) iron. b) zinc. c) copper. d) lead.

krok68 [10]

Explanation:

Copper.................

hope it's help you ❤️

5 0

3 years ago

Read 2 more answers

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

Ket [755]

Answer:- 10 L of ethane.

Solution:- The given balanced equation is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.

Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:

$14LO_2(\frac{2LC_2H_6}{7LO_2})$

= $4LC_2H_6$

From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L

Hence, 10 L of ethane will be remaining.

5 0

3 years ago