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ahrayia [7]
3 years ago
6

A 50.0-gram block of copper at 10.0°C is carefully lowered into 100.0 grams of water at 90.0°C in an insulated container. Which

statement describes the transfer of heat in this system?
(1) The water loses heat to the block until both are at 10.0°C.
(2) The block gains heat from the water until both are at 90.0°C.
(3) The water loses heat and the block gains heat until both are at the same temperature that is between 10.0°C and 90.0°C.
(4) The water gains heat and the block loses heat until both are at the same temperature that is between 10.0°C and 90.0°C.
Chemistry
1 answer:
Andrei [34K]3 years ago
5 0
First consider and take note of the following: 
1. When 2 or more finite amounts of substances are mix together (with different temperature) the final temperature is never equivalent to the initial temperature of any of the substances involved. 
2. The substances meet at an equilibrium temperature. 
3. Heat transfers from a higher temperature to a lower temperature. 

Only choice (3) satisfies the conditions. 
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Answer:

Both are similar concepts.

Sound is the vibration of air particles (compression and expansion) the can reach your ears. But you can have vibration being propagated in liquids and solids as well.

Some sounds are generated in structures, so the vibration of a structure is converted to sound in air — for instance, a loudspeaker.

Explanation:

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Jerod used a golf club to hit a ball what was the action and reaction
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The action was him hitting the ball the reaction was the ball moving after being hit
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3 years ago
Read 2 more answers
Can someone help me please? I’d appreciate it! Also please do not send me a link or delete your answer! Thanks
chubhunter [2.5K]

Answer:

14.53ML

Explanation:

V1=218

V2=?

P2=15p1

USING BOYLE'S LAW

P1V1=P2V2

V2=P1V1/P2=P1(218ML)/15P1

=14.53ML

5 0
3 years ago
28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cir
Damm [24]
<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

         = 0.0063 mole

<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

                             = 1.21 g citric acid

<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

   = (0.0050 × 0.25)÷ 0.30

   = 0.004167 L or

   = 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

8 0
3 years ago
What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium c
LuckyWell [14K]

Answer:

[A g + ]  =  3.12 *10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^− 12

Step 2: The balanced equation

2 A g + ( a q )  +  C rO4^2-  ( a q )  →Ag2CrO4 (  s)

Step 3:

[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 -0.088 = 0.123 M

1.2 * 10^− 12  = ( 2 x ) ²*( 0.123M + x )

[ C O 3^ −2] f  >>  x

1.2 * 10^− 12  = ( 2 x ) ²* 0.123M

x = 1.56 * 10^-6

[A g + ] f  = 2x = 3.12 *10^-6 M

,

4 0
3 years ago
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