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Helen [10]
2 years ago
5

(Help ASAP) in the summer many people in Argentina which is located in the southern hemisphere enjoy going to the beach which di

agram to the right shows the position of earth when it is summer in Argentina?

Chemistry
1 answer:
Studentka2010 [4]2 years ago
4 0

Answer:

I think it's B

Explanation:

A and C are out of the equation because Earth's axis isn't tilted on them and has the entirety of Argentina covered in shadow. The only answer left would be B.

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1. Naturally occurring europium (Eu) consists of two isotopes was a mass of 151 and 153. Europium- 151 has an abundance of 48.03
sergejj [24]
The atomic mass of Europium is 152 amu

Work:
151(0.4803) = 72.52 amu
153(0.5197) = 79.5 amu
72.5 + 79.5 = 152 amu
3 0
3 years ago
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What is the volume of a liquid that has a density 3.65g/ml and a mass of 5.61g
Lorico [155]

Answer:

<h2>1.54 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{5.61}{3.65}  \\  =1.536986...

We have the final answer as

<h3>1.54 mL</h3>

Hope this helps you

3 0
3 years ago
The rotational period of the moon is_____
Ivahew [28]

Answer:

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Explanation:

6 0
3 years ago
What type of substances is filtration useful for seperating? Give me an example.
Butoxors [25]
Filtration is useful for separating sediments that are carried in water for easy drinking without the risk of disease - an example of this is a water purifier
4 0
3 years ago
Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing
Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of AgNO_3 = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of AgNO_3 = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

7 0
3 years ago
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