Which temperature is equal to 120. K?
(1) -153°C (3) +293°C
(2) -120.°C (4) +393°C

Answers

Answer 1

Answer: First of all, the formula for finding Kelvin is Celsius + 273
Therefore, if we subtract 273, we get the temperature in degrees
120 - 273 = - 153

Therefore, the answer is (1), or -153 degrees Celsius

Hope this helped!! :D

Answer 2

Answer:

Final answer:

The temperature 120 K converts to -153°C. This conversion is made by subtracting 273.15 from the Kelvin temperature.

Explanation:

To convert Kelvin (K) to Celsius (°C), you subtract 273.15 from the Kelvin temperature. For example, 120 K becomes 120 - 273.15, equaling -153.15°C. Rounded, this is approximately -153°C. It's essential to note that 0 K is absolute zero, representing the lowest possible temperature, which corresponds to -273.15°C on the Celsius scale. Therefore, the correct answer is (1) -153°C. This conversion is fundamental in relating temperatures on the Kelvin and Celsius scales, where Celsius measures temperature relative to the freezing and boiling points of water, and Kelvin starts from absolute zero, the absence of thermal energy.

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How many grams of magnesium oxide can be produced when 97.2 g Mg react with 88.5 g O2?

Answers

First we establish the chemical reaction equation.
2Mg + O2 = 2MgO.
Then we need to know which of the reactant is limiting. We do that by multiplying the number of moles of reactant to the stoichiomeric coefficient. For Magnesium: 97.2 g $* (mol)/(24g) *2 = 8$
For O2: 88.5g* $(mol)/(32g) *1 = 2.7$
Since O2 has the smaller moles, this is the Limiting reactant. Then we solve basing on the number of O2 used.

88.5g $* (mol)/(32g) * (2)/(1) * (56g)/(mol) = 309.75 g$

Answer : The mass of magnesium oxide produced will be, 161.2 g

Solution : Given,

Mass of Mg = 97.2 g

Mass of $O_2$ = 88.5 g

Molar mass of Mg = 24.3 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of MgO = 40.3 g/mole

First we have to calculate the moles of Mg and $O_2$ .

$\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=(97.2g)/(24.3g/mole)=4\text{ moles}$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=(88.5g)/(32g/mole)=2.7\text{ moles}$

The balanced reaction is,

$2Mg+O_2\rightarrow 2MgO$

As, 2 moles of Mg react with 1 moles of $O_2$

So, 4 moles of Mg react with $(4)/(2)=2$ moles of $O_2$

From this we conclude that the $O_2$ is in excess amount and Mg is in limited amount.

Now from the reaction we conclude that

As, 2 moles of Mg react to give 2 moles of MgO

So, 4 moles of Mg react to give 4 moles of MgO

Now we have to calculate the mass of MgO.

$\text{Mass of MgO}=\text{Moles of MgO}* \text{Molar mass of MgO}$

$\text{Mass of MgO}=(4moles)* (40.3g/mole)=161.2g$

Therefore, the mass of magnesium oxide produced will be, 161.2 g

A solution of h2so4(aq) with a molal concentration of 6.80 has a density of 1.318 g/ml. what is the molar concentration of this solution?

Answers

The molalconcentration of the solution is 5.38 M.

The molal concentration of sulphuric acid = 6.80 m

The density of the solution = 1.318 g/ml

Mass of the sulphuric acid in the solution is,

$= 6.80 * 9.80$

$=666.4 \: g$

The mass of sulphuric acid present in the solution is 666.4 g.

Let the mass of the solvent in which sulphuric acid is dissolved be 1000 grams.

So the total mass of the solution is,

= Mass of solvent + Mass of sulphuric acid

$= 666.4 \: + \: 1000$

$= 1666.4 \: g$

The total volume of the solution is,

$= (1666.4 * 1)/(1.318 * 1000)$

$= 1.264 \: L$

The volume of the solution is 1.264 liters.

The molal concentration of this solution is,

$= (6.80)/(1.264)$

$= 5.38 \: M$

Therefore, the molal concentration of the solution is 5.38 M.

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WHAT IS CLEVAGE PLANE

Answers

Cleavage, in mineralogy, is the tendency of crystalline materials to split along definite crystallographic structural planes.

How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mass solution? Show all of the work needed to solve this problem.

Answers

M1m1 = M2m2

where M1 is the concentration of the stock solution, m1 is the mass of the stock solution, M2 is the concentration of the new solution and m2 is its new mass.

M1m1 = M2m2

.925(m1) = .35(250)

m1 = 94.59 g

Answer: 94.59 grams of 92.5 % $H_2SO_4$ by mass solution will be needed.

Explanation:

Mass of sulfuric acid is 250 grams of 35 % by mass solution:

$35=(x)/(250 g)* 100$

$x=87.5 g$

Mass of $H_2SO_4$ in 250 g of 35 % solution = 87.5 g

Mass of 92.5 % $H_2SO_4$ needed to make 35 % by mass solution.

$92.5=\frac{87.5 g}{\text{mass of the solution required}}* 100$

Mass of the solution required = 94.59 g

94.59 grams of 92.5 % $H_2SO_4$ by mass solution will be needed.

Which of the following materials is likely to be the best insulator?-silver
-phosphorus
-silicon
-calcium

Answers

Answer: Option (b) is the correct answer.

Explanation:

Insulator are the substances which do not conduct electricity. It is known that metals are good conductors of electricity. Whereas non-metals do not conduct electricity.

Out of the given options, silver, and calcium is a metal that means they are good conductors of electricity and are not insulators.

On the other hand, silicon is a metalloid, that is, it shows properties of both metals and non-metals. Hence, it will also show conductivity.

Phosphorus is a non-metal and therefore, it does not conduct electricity. As a result, out f the given options it will act as the best insulator.

The element most likely to be the best insulator is the worst conductor among the four when you test for their conductivity. The answer among the four options given is 'phosphorus'. I hope this helps you on your assignment.

Why are protons in an atom so important? Question 5 options: A. Protons determine the atoms' identity (type of element) and where it is located on the periodic table. B. Protons are bigger than all the other particles in an atom. C. Protons determine an atom's chemical properties and how it will react with other elements. D. There are more protons than electrons or neutrons in a neutral atom.