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lora16 [44]
3 years ago
12

If 30.0 g absorbs 255 J of heat, how much does the temperature of the water change? Would the temperature be increasing or decre

asing?
Chemistry
1 answer:
Nutka1998 [239]3 years ago
3 0
<h3>Answer:</h3>

Change in temperature = 2.03°C, the temperature is increasing

<h3>Explanation:</h3>
  • To calculate the quantity of heat absorbed or released by a substance we multiply mass of the substance by it's specific heat capacity and the temperature change.
  • Therefore, Quantity of heat, Q = mass × specific heat × change in temperature

In this case;

Mass of water = 30.0 g

Quantity of heat absorbed = 255 J

Specific heat capacity of water = 4.186 J/g°C

Rearranging the formula, Δt = Q ÷ mc

Δt = 255 J ÷ (4.186 J/g°C×30.0 g )

  = 2.03 °C

The temperature change is 2.03°C, the temperature is therefore increasing.

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Salts of Transition elements.

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In an oxidation-reduction reaction, 0.0450 mol of aqueous FeSO4 (source of Fe2+) reacts completely with 120.0 mL of an acidified
igor_vitrenko [27]

Answer:

0.075 M

Explanation:

5Fe²⁺(aq) + MnO₄⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(ℓ)

Using the moles of Fe²⁺ that reacted, we can <em>calculate the reacting moles of MnO₄⁻</em>:

0.0450 mol Fe⁺² * \frac{1molMnO_{4}^{-}}{5molFe^{+2}} = 0.0090 mol MnO₄⁻

Now we divide the moles of MnO₄⁻ by the volume in order to <u>calculate the molarity of the solution</u>, keeping in mind that 120.0 mL = 0.120 L.

0.0090 mol MnO₄⁻ / 0.120 L = 0.075 M

7 0
4 years ago
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In a chemical process, three bottles of a standard fluid are emptied into a larger container. A study of the individual bottles
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Answer:

a

The expected value is = 39 units

The standard deviation is = 0.1212 unites

b

The probability is = 0.2047 units

Explanation:

The explanation of this answer is shown on the first uploaded image

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3 years ago
Experiments in which electricity was discharged into a vessel containing hydrogen gas, water vapor, methane, and ammonia were co
alexdok [17]

Answer:

Stanley Miller and Harold Urey

Explanation:

Experiments in which electricity was discharged into a vessel containing hydrogen gas, water vapor, methane, and ammonia were conducted by Miller and Urey.

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Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?
erastova [34]

Answer:

A precipitate will form, BaCO₃

Explanation:

When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:

Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]

<em>Where the concentrations of the ions are the concentrations in equilibrium</em>

<em />

For actual concentrations of a solution, you can define Q, <em>reaction quotient, </em>as:

Q = [Ba²⁺] [CO₃²⁻]

<em>If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form</em>.

Actual concentrations of Ba²⁺ and CO₃²⁻ are:

[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M

[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M

<em>100.0mL is the volume of the mixture of the solutions</em>

<em />

Replacing in Q expression:

Q = [Ba²⁺] [CO₃²⁻]

Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]

Q = 1.5x10⁻⁷

As Q > Ksp

<h3>A precipitate will form, BaCO₃</h3>

<em />

8 0
3 years ago
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