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lawyer [7]
3 years ago
8

Using the picture below, once air mass A reaches location B, the weather conditions at location B will most likely become

Chemistry
1 answer:
olga2289 [7]3 years ago
3 0
I think the answer is a
You might be interested in
How much heat is required to melt 20 g of gold at 1064.18 °C with a heat of fusion of 64 J/g *
andrew-mc [135]

Answer:

1280J are required.

Explanation:

Heat of fusion is defined as the amount of heat required to change its state from liquid to solid at its melting point at constant pressure.

As heat of fusion of gold is 64J/g, there are required 64J to melt 1g of gold at its melting point. The energy required to melt 20g is:

20g * (64J/g) =

1280J are required

6 0
3 years ago
The major dissolved constituents of ocean water, such as sodium and chloride ions, are __________ properties of seawater.
Darya [45]

Answer:

Conservative Properties of Seawater

Explanation:

The Conservative properties of seawater refer to those properties that cannot be altered due to the occurrence of physical, chemical and biological processes, over the large oceanic bodies. This typically comprises properties such as the temperature and also there is a high concentration of both sodium and chloride ions, which increases the salinity of the oceans.

These conservative properties occur in almost a fixed amount, or it most probably changes at a very slower rate through time. They can be considered to have a long residence time.

5 0
4 years ago
Help please I got the back but I’m struggling with the front
musickatia [10]
41. Mercury,Venus, Earth,mars,Jupiter, Saturn,Uranus and Neptune
42. Nebula - Their birth places are huge, cold clouds of gas and dust
8 0
3 years ago
How many moles of gas would occupy 22.4 L at 273K in one arm?
andrew-mc [135]

Answer:

1 mol

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the provided information in the question;

V = 22.4L

T = 273K

P = 1 atm

R = 0.0821 Latm/molK

n = ?

Using PV = nRT

n = PV/RT

n = (1 × 22.4) ÷ (0.0821 × 273)

n = 22.4 ÷ 22.4

n = 1mol

4 0
3 years ago
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
Eva8 [605]

Answer:

m_{Ag}=135.8gAg

Y=88.4\%

Explanation:

Hello there!

In this case, according to the described chemical reaction, it is possible to compute the theoretical mass of silver as mass via the 1:2 mole ratio of copper to silver and their atomic mass in the periodic table, in order to perform the following stoichiometric setup:

m_{Ag}=40.gCu*\frac{1molCu}{63.55gCu}*\frac{2molAg}{1molCu}*\frac{107.87gAg}{1molAg}\\\\   m_{Ag}=135.8gAg

Next, given the actual yield of 120 g, we compute the percent yield via:

Y=\frac{120g}{135.8g}*100\%\\\\Y=88.4\%

Regards!

4 0
3 years ago
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