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prisoha [69]
2 years ago
10

A steel plaque is attached to a block of wood. The steel and the wood have the same mass. After the steel plaque and the wooden

block were left outside in the sun for a period of time, both the steel and the wood had increased in temperature, but the steel had a higher temperature than the wood.
It’s possible that the steel plaque had a higher temperature than the wooden block because it________ its surroundings.
If the steel and the wood received the same amount of energy from the Sun, then the specific heat capacity of the steel must be_______ the specific heat capacity of the wood.

choices for blank 1:absorbed more energy from --released more energy to--way better insulated from

choices for blank 2: equal to --less then---greater then
Chemistry
2 answers:
Alex2 years ago
8 0

General Remark

The amount of energy transferred from the sun to the wood and steel is determined by m*c*deltaT. [So far, you really only have 2 choices for heat energy transfer. One is the one you are using m*c*deltaT and the other one is Mass * heat. This latter is for phase changes].

Solution

What causes the temperature to be higher in one part of the sample than on the other part? If they both receive the same amount of heat (as in getting their energy from the sun) then why does one deltaT exceed the other? Look around at the parts of the formula. That's very important in science. The mass (m) is not going to affect the different temperatures. The mass is the same in both cases.

It won't be deltaT either. DeltaT has already done it's job. It's recorded that there is a temperature difference. Steel has a larger deltaT than the wood.

What's left?

It must be the c. What are the units on c? If you check it out it is joules / (oC * grams) or something that means grams and deltaT. So c must be lower than steel than for wood. Why? Because it takes less energy to heat up the steel. Small c means large deltaT if the temperature is greater in the iron.

Answers

<em><u>Blank One </u></em>absorbed more energy from. c is small.

<em><u>Blank Two</u></em> Less than.

This is an extremely important question to get a handle on. More and Less are very important key words in Science.


Nina [5.8K]2 years ago
6 0

Answer:

<u>Blank One</u> = Absorbed more energy from

<u>Blank Two</u> = Less than

Explanation:

You might be interested in
) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H
Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

pH = -log([H_{3}O^{+}])

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

<u>Trimethyl ammonium</u>:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O  →  (CH₃)₃N + H₃O⁺    

1.0 - x                               x           x  

Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}

10^{-pKa} = \frac{x*x}{1.0 - x}

10^{-9.80}(1.0 - x) - x^{2} = 0    

By solving the above equation for x we have:  

x = 0.097 = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01                                      

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x                        x             x

Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}

10^{-10} = \frac{x^{2}}{1.0 - x}

1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0
2 years ago
How many hydrogen molecules are there in 1 ton of hydrogen?​
zaharov [31]
Hydrogen gas(H2) has a molar mass of 2 g. Molar mass of a substance is defined as the mass of 1 mole of that substance. And by 1 mole it is meant a collection of 6.022*10^23 particles of that substance.

So number of moles of H2 are 0.5 in this case. And thus it means there are (6.022*10^23)*0.5 particles( here they are molecules) in 1g of H2.
6 0
2 years ago
Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K
Dmitry [639]

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

5 0
2 years ago
A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained
Tamiku [17]

Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow  Na_2SO_4(aq) + CO_2(g) + H_2O(l)

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

5 0
2 years ago
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
VLD [36.1K]

Answer:- 64015 J

Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.

density of water is 1 g per mL.

So, the mass of water = 4250mL(\frac{1g}{1mL})  = 4250 g

Final temperature of water after adding the hot copper bar to it is 26.15 degree C.

So, \Delta T for water = 26.15 - 22.55 = 3.60 degree C

Specific heat for water is 4.184 \frac{J}{g.^0C}

The heat gained by water is calculated by using the formula:

q=mc\Delta T

where, q is the heat energy, m is mass and c is specific heat.

Let's plug in the values in the formula and do the calculations:

q=4250g*\frac{4.184J}{g.^0C}*3.60^0C

q = 64015 J

So, 64015 J of heat is gained by the water.



5 0
2 years ago
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