Answer the questions below:

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

3 years ago

What are two ways carbon returns from animal into the water

soldi70 [24.7K]

Split and merge into it. While they are alive, carbon returns from animals into water through waste products from respiration and defecation/urination. Another way when they are dead is from decaying remains. While they are alive, carbon returns from animals into water through waste products from respiration and defecation/urination.

Good enough?

8 0

3 years ago

what is the balanced equation of magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide

fenix001 [56]

Answer:

ExplaAt high temperatures MgCO3 decomposes to magnesium oxide and carbon dioxide. This process is important in the production of magnesium oxide. This process is called calcining: MgCO3 → MgO + CO2 (ΔH = +118 kJ/mol)

nation:

6 0

3 years ago

Please help me!!!30 pts

Tanya [424]

Answer:

Question 7 to 12 are given in attached file because character limit is only 5000

Explanation:

1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of sulfur. Calculate the empirical formula of this compound.

Given data:

Mass of sample = 15 g

Mass of sodium = 8.83 g

Mass of sulfur = 6.17 g

Empirical formula = ?

Solution:

Number of gram atoms of Na = 8.83 / 23 = 0.4

Number of gram atoms of S = 6.17 / 32 = 0.2

Atomic ratio:

Na : S

0.4/0.2 : 0.2/0.2

2 : 1

Na : S = 2 : 1

Empirical formula is Na₂S.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Given data:

Mass of phosphorus = 4.433 g

Mass of oxygen = 10.150 g - 4.433 g = 5.717 g

Empirical formula = ?

Solution:

Number of gram atoms of P = 4.433 / 30.9738 = 0.1431

Number of gram atoms of O = 5.717/ 15.999 = 0.3573

Atomic ratio:

P : O

0.1431/0.1431 : 0.3573/0.1431

1 : 2.5

P : O = 2(1 : 2.5)

Empirical formula is P₂O₅.

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Given data:

Percentage of sodium = 36.48%

Percentage of sulfur = 25.41%

Percentage of oxygen = 38.11%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 36.48 / 23 = 1.6

Number of gram atoms of S = 25.41/ 32 = 0.8

Number of gram atoms of O = 38.11/ 16 = 2.4

Atomic ratio:

Na : S : O

1.6/0.8 : 0.8/0.8 : 2.4/0.8

2 : 1 : 3

Na: S : O = 2 : 1 : 3

Empirical formula is Na₂SO₃.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Given data:

Percentage of iron = 63.52%

Percentage of sulfur = 36.48%

Empirical formula = ?

Solution:

Number of gram atoms of Fe = 63.52 / 55.845 = 1.14

Number of gram atoms of S = 36.48 / 32 = 1.14

Atomic ratio:

Fe : S

1.14/1.14 : 1.14/1.14

1 : 1

Fe : S = 1 : 1

Empirical formula is FeS.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 % oxygen. Find the empirical formula of this compound.

Given data:

Percentage of sodium = 32.38%

Percentage of sulfur = 22.65%

Percentage of oxygen = 44.99%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 32.38 / 23 = 1.4

Number of gram atoms of S = 22.65/ 32 = 0.7

Number of gram atoms of O = 44.99/ 16 = 2.8

Atomic ratio:

Na : S : O

1.4/0.7 : 0.7/0.7 : 2.8/0.7

2 : 1 : 4

Na: S : O = 2 : 1 : 4

Empirical formula is Na₂SO₄.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that 4.00 grams of calcium are present. What is the empirical formula of the compound formed?

Given data:

Mass of sample = 20g

Mass of bromine = 20 g - 4 g = 16 g

Mass of calcium = 4 g

Empirical formula = ?

Solution:

Number of gram atoms of bromine = 16 / 80= 0.2

Number of gram atoms of calcium = 4/ 40= 0.1

Atomic ratio:

Ca : Br

0.1/0.1 : 0.2/0.1

1 : 2

Ca: Br = 1 : 2

Empirical formula is CaBr₂.

5 0

3 years ago

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Identify the two types of radiation shown in the figure below. Explain your reasoning.

saul85 [17]

Answer:- $\gamma$ and $\alpha$ radiation.

Explanations:- Alpha particle is helium nucleus with two protons and two neutrons and so it has +2 charge where as gamma is a form of light, so these are rays and not particles. Since these are rays, the charges for gamma rays is zero means they are neutral.

When alpha particles are passed through electromagnetic field they are repelled by the positive plate and attracted by the negative plate as there is an attraction between opposite charges. So, the line we see towards positive plat is representing an alpha particle radiation.

Since gamma rays are neutral, they are not deflected in electromagnetic field and go straight. So, the other line in the diagram is representing gamma rays radiation.

5 0

3 years ago

Read 2 more answers