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White raven [17]
2 years ago
9

1-Two metal samples with equal masses are heated by equal amounts of energy. The temperature of Sample A rose 10oC while the tem

perature of Sample B increased by 20oC. What can you determine about the specific heat capacity of these substances?
2-A sample of a metal with a mass of 10.0 g cooled from 50.4 oC to 25.0 oC, releasing of 168 Joules of heat. What is the specific heat of the metal sample?
Chemistry
1 answer:
makkiz [27]2 years ago
8 0

Answer:

the channel tunnel has been a very busy and challenging experience and we are mot doing.

Explanation:

ok and I like the country studies opportunity and I will be found by him too as he is a fact in the sentence.

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What is the percent by volume of a solution formed by mixing 15 mL of hydrochloric acid into water, forming 75 mL of solution?
timama [110]

Answer:

= 17% (v/v)  

Explanation:

% by Volume = (Volume HCl / Volume of Solution)100%

= 15ml/(75ml + 15ml) x 100% = 16.66666667% (calculator answer)

= 17% (v/v)  2 sig figs. based on volume values given.

4 0
3 years ago
What does it mean if a solution is a "weak acid "? Strong acid? Give logical ph values for each
galina1969 [7]

Answer:

A weak acid is an acid that partially dissociates into its ions in an aqueous solution or water. In contrast, a strong acid fully dissociates into its ions in water. At the same concentration, weak acids have a higher pH value than strong acids.

Explanation:

:)

5 0
2 years ago
I’ll mark brainliest please
DaniilM [7]
I’m assuming your just writing the formula? If so
Potassium chloride: KCL
Potassium nitride: KNO2
Potassium sulfide: K2S
calcium chloride: CaCl2
Calcium nitride: Ca3N2
Calcium sulfide: CaS
Silver chloride: AgCl
Silver nitride: Ag3N
Silver sulfide: Ag2S
Manganese (||) chloride: MnCl2
Manganese (||) nitride: Mn3N2
Manganese (||) sulfide: MnS
5 0
2 years ago
A hand-operated conduit bender with a shoe for bending 1-inch EMT can also be used to bend
mixer [17]
If your choices are the following:
A. 1-inch IMC.
B. 1-inch rigid conduit.
C. 3/4 inch IMC.
D. 1/2 inch EMT.

Then the answer is C.
6 0
3 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
2 years ago
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