An iron bar at 325 K is placed in a sample of water. The iron bar gains energy from the water if the temperature of the water is(1) 65 K (3) 65°C
(2) 45 K (4) 45°C

Answers

Answer 1
Answer: 0K = -273 C
So
273K = 0C
So
325K = 52C
So
The water would need to be warmer than 52C to give the iron bar energy
So the answer is (3) 65C
Answer 2
Answer:

Final answer:

The iron bar at 325 K will gain energy from the water if the water temperature is higher, which is the case when it is 65°C (338 K). The bar would lose energy to the other given temperatures (45 K, 65 K, 45°C).

Explanation:

The transfer of thermal energy, or heat, is from a body at higher temperature to a body at lower temperature. The iron bar in your question is at 325 K. The temperatures of the water samples are closer to room temperature (around 293 K) when expressed in Kelvin. Therefore, the iron bar will gain energy from the water if the temperature of the water is 65°C (338 K), as it is the only option above the temperature of the iron bar. It would lose energy to the other options (45 K, 65 K, 45°C).

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A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure. How many grams of N2 must react to form 1.7 grams of ammonia, NH3? 0.0058 g .052 g 1.4 g 2.8 g

Answers

Since the percent yield of the reaction is 98%, that means that 1.7g of ammonia is 98% of the theoretical yield so you need to find what the theoretical yield is by dividing 1.7 by .98 to get 1.735g.  Since you know that the theoretical yield is 1.735g you can use stoichiometry to figure out how much N₂ is required in order to produce 1.735g of ammonia.  

To do that you first need to find the balanced chemical equation for the reaction which is N₂+3H₂⇒2NH₃.  Then you need to find how many moles of ammonia are in 1.735g by dividing 1.735 by the molar mass of ammonia (17g/mol) to get 0.1021mol of ammonia.  After that you can find the number of moles of N₂ required to make 0.1021mol of ammonia by using the fact that 1mol N₂ is used to make 2mol NH₃ so you divide 0.1021 by 2 to get 0.05103mol N₂.  lastly you can find the number of grams of N₂ by multiplying 0.05103 by the molar mass of N₂ (28g/mol) to get 1.42g of N₂.  Therefore 1.4g of N₂ is required to make 1.7g of ammonia.

I hope this helps. Let me know in the comments if anything is unclear.

A muffin has a mass of 100g and a volume of 500cm^3. What is the density of the muffin

Answers

Answer:

d = 0.2 g/cm³

General Formulas and Concepts:

Density = mass / volume

Explanation:

Step 1: Define

m = 100 g

V = 500 cm³

d = ?

Step 2: Find density

  1. Substitute:                              d = 100 g/500 cm³
  2. Evaluate:                                 d = 1/5 g/cm³
  3. Evaluate:                                 d = 0.2 g/cm³

All solids have an orderly internal structure with repeating bonding patterns between atoms, ions, or molecules. a. True
b. False

Answers

the answer is false cause i tried true first and got it wrong.

Answer:

False

Explanation:

Not all solids have an orderly internal structure, although some do which are called crystalline solids. Some examples are metals and table salt. Solids without a orderly internal structure are called amorphous solids. Examples of this are glass, plastics, and rubber.

The ratio of oxygen to carbon by mass in carbon monoxide is 1.33:1.00. part a find the formula of an oxide of carbon in which the ratio by mass of oxygen to carbon is 2.00:1.00.

Answers

To calculate the molecular formula, convert the mass ratio into molar ratio as follows:

mass ratio of O:C=2:1

molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus, number of moles can be calculated as follows:

n=\frac{m}{M}

calculating ratio,

O:C=\frac{2 g}{16 g/mol}:\frac{1 g}{12 g/mol}=\frac{1}{8}:\frac{1}{12}=12:8=3:2

thus, molecular formula will be C_{2}O_{3}

Final answer:

If the ratio of oxygen to carbon by mass is 2.00:1.00, the formula of the oxide of carbon is CO₂ (carbon dioxide). This is because CO₂ has twice as much oxygen per amount of carbon as compared to CO (carbon monoxide). This situation adheres to the law of multiple proportions.

Explanation:

This question revolves around the concept of stoichiometry in chemistry, particularly in relation to the law of multiple proportions and finding the formula of an oxide of carbon given specific mass ratios.

Firstly, in carbon monoxide (CO), the ratio of oxygen to carbon by mass is 1.33:1.00.

However, when the ratio of oxygen to carbon by mass increases to 2.00:1.00, we are now dealing with a compound that contains twice as much oxygen per amount of carbon. In essence, this would be carbon dioxide (CO₂). The mass ratio of oxygen to carbon in CO₂ is indeed 2:1 (32 g/mol oxygen: 12 g/mol carbon).

This situation illustrates the law of multiple proportions - in this case, the two oxides of carbon (CO and CO₂) contain elements combined in ratios of small whole numbers.

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How many atoms are in 0.750 moles of zinc?

Answers

There are approximately 4.52 x 10^23 atoms in 0.750 moles of zinc.

How to find the number of atoms

To determine the number of atoms in a given amount of a substance, you can use Avogadro's number, which is approximately 6.022 x 10^23 atoms/mol.

Given that you have 0.750 moles of zinc, you can calculate the number of atoms using the following steps:

Multiply the given number of moles by Avogadro's number:

0.750 moles * (6.022 x 10^23 atoms/mol) = 4.5165 x 10^23 atoms

Round the result to an appropriate number of significant figures:

Since the value given has three significant figures, the final answer should be rounded accordingly. Therefore, the number of atoms in 0.750 moles of zinc is approximately 4.52 x 10^23 atoms.

So, there are approximately 4.52 x 10^23 atoms in 0.750 moles of zinc.

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(doesn't really matter that it's zinc :) )
It is 0.75*N_A, where N_A=6.023*10^(23) (in some books 6.022).

Using the balanced equation below, how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride?CSF + XeF6 → CsXeF7

Answers

Answer:

27.9 g

Explanation:

CsF + XeF₆ → CsXeF₇

First we convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles, using its molar mass:

  • Molar mass of CsXeF₇ = 397.193 g/mol
  • 73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇

As 1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.

Now we convert 0.184 moles of CsF to moles, using the molar mass of CsF:

  • Molar mass of CsF = 151.9 g/mol
  • 0.184 mol * 151.9 g/mol = 27.9 g