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Mumz [18]
3 years ago
5

Mn or Co is treated with 1 mL of 6 M HNO3 followed immediately by the drop-wise addition of 3% hydrogen peroxide until the preci

pitate fully dissolves. The test tube is then heated in a water bath until bubbling ceases. At this stage of the experiment, which elements could be found in the resulting LIQUID?
Chemistry
1 answer:
ki77a [65]3 years ago
3 0

Answer:

Mn, Co,  H, N and O

Explanation:

When metallic Mn or Co reacts with concentrated HNO₃, they dissolve completely in the acid . The products formed are the nitrates of the metals, water and a mixture of NO and NO₂ gases. Hydrogen peroxide is added to oxidize any NO to NO₂ which is less harmful.

The equations of the reaction are given below;

Mn + HNO₃ ====> Mn(NO₃)₂ + NO + H₂O

Co + HNO₃ ====> Co(NO₃)₂ + NO + H₂O

H₂O₂ + NO ====> H₂O + NO₂

The resulting solution will be the soluble nitrates of manganese and cobalt as well water and this solution contains the following elements; Mn, Co, N, H And O.

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For the following reaction in aqueous solution, identify all those species that will be spectator ions. Select all that apply.
FromTheMoon [43]

The reaction is incorrect. The correct reaction is: Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃. The options are:

A. Hg₂SO₄

B. Na₂SO₄

C. Na⁺

D. NO₃⁻

E. SO₄⁻²

F. Hg₂(NO₃)₂

G. NaNO₃

H. Hg⁺²

Answer:

C, D, E, H

Explanation:

The ions are formed after dissociation of the compound in the solution. When they're negatively charged, they're called anions, when they're negatively charged, they're called cations. If the ion is presented on both sides of the reaction, it is called a spectator ion.

Thus, the reaction given:

Na₂SO₄ + Hg(NO₃)₂ → HgSO₄ + 2NaNO₃

So, let's do the dissociation.

Na₂SO₄ is formed by the ions Na⁺ and SO₄⁻²;

Hg(NO₃)₂ is formed by the ions Hg⁺² and NO₃⁻;

HgSO₄ is formed by the ions Hg⁺² and SO₄⁻²;

NaNO₃ is formed by the ions Na⁺ and NO₃⁻.

Thus, the spectator ions are Na⁺, SO₄⁻², Hg⁺², and NO₃⁻.

7 0
3 years ago
Boyle's Law relates temperature to gas volume and Charles's Law relates pressure to gas volume.
vagabundo [1.1K]

False100% sure of it

4 0
3 years ago
Read 2 more answers
In the molecule bri, which atom is the negative pole?
Harrizon [31]
<span>BrI molecule is made of Bromine and Iodine. The polarity depends on the electronegativity of Bromine and Iodine. The one which has more electronegativity value shall be the negative pole Br: 2.96 I: 2.66 Hence the negative pole shall be Bromine.</span>
5 0
3 years ago
1. Three
Bess [88]

Answer:

Explination:

Given Data:

                  Trail 1          Trial 2         Trial 3

Student A 448.0 cm 485.6 cm 463.4 cm

Student B 450.5 cm 441.3 cm         446.8 cm

Student C 422.6 cm 445.2 cm 432.7 cm

Accepted Value = 435.0 cm

Required:

A: Accurate measurement =?

B: Reason for the answer =?

C: Precise measurement =?

D: Reason for the answer =?

Solution:

Student A:

Trail 1: 435.0 cm – 448.0 cm = (13.0 cm greater than accepted value)

Trail 2: 435.0 cm – 485.6 cm = (50.6 cm greater than accepted value)

Trial 3: 435.0 cm – 463.4 cm = (28.4 cm greater than accepted value)

Student B:

Trail 1: 435.0 cm – 450.5 cm = (14.5 cm greater than accepted value)

Trail 2: 435.0 cm – 441.3 cm = (6.3 cm greater than accepted value)

Trial 3: 435.0 cm – 446.8 cm = (11.8 cm greater than accepted value)

Student C:

Trial 1: 435.0 cm – 422.6 cm = (12.4 cm less than accepted value)

Trial 2: 435.0 cm – 445.2 cm = (10.2 cm greater than accepted value)

Trial 3: 435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

A: The 3rd trial of the student C is accurate measurement = 432.7 cm.

B: The 3rd value of students’ C measurement is accurate because it is quite near to the accepted value, i.e.  435.0 cm.

As we know that Accuracy refers to the closeness of a measured value to a standard or known value.

This value has only the difference of 1.3 cm.

435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

All the other have larger difference which is as above.

___________________

Student A:

1. 448.0 cm – 485.6 cm = (37.6 cm far)

2. 485.6 cm – 463.4cm = (22.2 cm far)

Student B:

1. 450.5 cm – 441.3 cm = (9.2 cm far)

2. 441.3 cm – 446.8 cm = (5.5 cm far)

Student C:

1. 422.6 cm – 445.2 cm = (22.6 cm far)

2. 445.2 cm – 432.7cm = (12.5 cm far)

So,

C: The values of Student B are more precise.

D: As we know that Precision refers to the closeness of two or more measurements to each other.

The measurements of student C are more close to each other. The values are only 9.2 cm and 5.5 cm far from each other.

5 0
3 years ago
What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0
2 years ago
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