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Xelga [282]
2 years ago
9

Substance A with a specific heat of 1.25 J/g∙oC and Substance B with a specific heat of 12.5 J/g∙oC. If both were exposed to the

same amount of heat which one would rise in temperature the most?
Chemistry
1 answer:
olga55 [171]2 years ago
5 0

Answer:

Substance A

Explanation:

Specific heat is defined by the amount of heat needed to raise the temperature of 1 gram of a substance 1 degree Celsius (°C).  

A higher specific heat means it takes more energy to increase the temperature of the substance compared to other substances. Since Substance A has a lower specific heat, it takes less energy to raise the temperature so it will rise in temp faster.

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a container holds 7.4 moles of gas hydrogen gas makes up 25% of the total moles in the container if the pressure is 1.24 atm wha
Tema [17]
This question is based on Dalton's Law of Partial Pressure which states that "the total pressure of a system of gas is equal to the sum of the pressure of each individual gas (partial pressure).

Now, Partial Pressure of a gas = (mole fraction) × (total pressure)

⇒ Partial Pressure of Hydrogen = \frac{1}{4}   ×   \frac{1.24}{1}
                  
                                                     =   0.31 atm

        Thus the Partial Pressure of Hydrogen in the container is 0.31 atm.
6 0
3 years ago
In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}

10=\frac{4}{3}*pi*{r_{L}}^{3}

{r_{L}}^{3}=\frac{7.5}{pi}

r_{L}=1.337cm

Step 2: Determine surface area of large sphere

A_{L}=4*pi*{r_{L}}^{2}

A_{L}=4*pi*1.337^{2}

A_{L}=22.463 cm^{2}

Step 3: Determine radius of small sphere

V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}

1.25=\frac{4}{3}*pi*{r_{s}}^{3}

{r_{s}}^{3}=\frac{0.9375}{pi}

r_{s}=0.668cm

Step 4: Determine surface area of small sphere

A_{s}=4*pi*{r_{s}}^{2}

A_{s}=4*pi*0.668^{2}

A_{s}=5.607 cm^{2}

Step 5: Determine total surface area of 8 small spheres

A_{S}=8*A_{s}

A_{S}=8*5.607

A_{S}=44.856 cm^{2}

A_{L}=22.463 cm^{2} - Surface area of 1 large sphere

A_{S}=44.856 cm^{2} - Surface area of 8 small spheres

Options:

  1. The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
  2. The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
  3. The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
  4. The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
  5. The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
  6. The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm
7 0
3 years ago
Read 2 more answers
Which of these is a mixture? a.water b.sugar c.carbon dioxide d.popcorn and salt​
Dmitry_Shevchenko [17]

Popcorn and salt as the mixtures are substances which are formed by mixing other substances

7 0
3 years ago
What is the definition of light in one sentence
bulgar [2K]

The dictionary definiton of light is:

the natural agent that stimulates sight and makes things visible.

I'll paraphrase/give a scientific definition:

Light is made of subatomic particles called photons, which jumps electron shells when they're excited, effecting the electromagnetic spectrum, creating visible light.

Hope this helps. Might not as I'm not perfect on this topic yet.

4 0
3 years ago
What volume does 2.25g of nitrogen gas, N2, occupy at 273 Celsius and 1.02 atm​
kotykmax [81]
<h2><u>Answer:</u></h2>

0.126 Liters

<h2><u>Explanation:</u></h2>

V = mRT / mmP

First, convert the 2.25g of Nitrogen gas into moles. (m in the equation above)

2.25g x 1 mole / 28.0g = 0.08036 moles = m

28.0g = mm

Next, convert the 273 Celsius into Kelvin. (T in the equation above)

273 Celsius + 273.15 = 546.15K = T

R = 0.08206L*atm/mol*K

(Quick Note: The R changes depending on the Pressure Unit so do not use this number every time.)

Now, plug everything into the equation.

V = (0.08036)(0.08206)(546.15)/(28.0)(1.02)

V = 0.126 L

5 0
3 years ago
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