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3 years ago

2. How much energy must be removed from a 50.0 g sample of steam in order to get it to condense

Chemistry

1 answer:

Strike441 [17]3 years ago

5 0

Answer:

113 kJ or 113,000J

Explanation:

Since we are condensing it, a physical change in the state of matter is occurring. 113 kJ or 113,000J because of the following equation: Q = 2260 times 50. 2260 is the Heat of Vaporization constant.

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Anyone know the answer to this one?

sergey [27]

Answer:

C) 0.24 M

Explanation:

The given chemical reaction is presented as follows;

H₂SO₄(aq) + 2 KOH(aq) → K₂SO₄(aq) + 2 H₂O(l)

The titration experiment results are;

Volume of H₂SO₄(aq) used = 12.0 mL

Volume of KOH (aq) used = 36.0 mL

Concentration of KOH (aq) = 0.16 M

The number of moles of KOH present, n = 0.16 M × 36/1000 = 0.00576 moles

From the given reaction, 1 mole of H₂SO₄ reacts with 2 moles of KOH to give 1 mole of K₂SO₄ and 2 moles of H₂O

Therefore, 0.00576 moles of KOH reacts with (1/2) × 0.00576 moles = 0.00288 moles of H₂SO₄

Therefore, for the reaction;

The number of moles of H₂SO₄ in 12.0 mL of H₂SO₄ = 0.00288 moles

The concentration of the H₂SO₄ = 0.00288 M/(12.0 mL) = 0.24 M

The concentration of H₂SO₄ in the reaction = 0.24 M.

5 0

3 years ago

What is true for a substance that releases energy

hammer [34]

The sun is a true substance that gives off energy.

6 0

4 years ago

Write a balanced equation for the combustion of C7H16(l) (heptane) -- i.e. its reaction with O2(g) forming the products CO2(g) a

JulsSmile [24]

Answer:

The standard enthalpy of reaction = -4854.7kJ

The difference: ΔH-ΔE = Δ(PV) = Δn.R.T = 9910.288 J ≈ 9.91 kJ

Explanation:

The balanced chemical equation for the combustion of heptane:

C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)

Given: The standard enthalpy of formation ( $\Delta H _{f}^{\circ }$ ) for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol

To calculate the standard enthalpy of reaction ( $\Delta H _{r}^{\circ }$ ) can be calculated by the Hess's law:

$\Delta H _{r}^{\circ } = \left [\sum \nu \cdot\Delta H _{f}^{\circ }(products) \right ] - \left [\sum \nu\cdot\Delta H _{f}^{\circ }(reactants) \right ]$

Here, $\nu$ is the stoichiometric coefficient

⇒ $\Delta H _{r}^{\circ } =$

$\left [ 7\times \Delta H _{f}^{\circ }\left (CO_{2}\right )+ 8\times \Delta H _{f}^{\circ }\left (H_{2}O \right )\right ]$

$- \left [1\times \Delta H _{f}^{\circ }\left (C_{7}H_{16}\right ) +11\times \Delta H _{f}^{\circ }\left (O_{2} \right ) \right ]$

$=\left [ 7\times \left (-393.5 kJ/mol \right )+ 8\times \left (-286 kJ/mol \right )\right ]$

$-\left [1\times \left (-187.8 kJ/mol \right ) +11\times \left (0 kJ/mol \right ) \right ]$

⇒ $\Delta H _{r}^{\circ } = \left [ \left (-2754.5 \right )+ \left (-2288 \right )\right ]\left -[ \left (-187.8 \right ) +\left (0 \right )\right ]$

⇒ $\Delta H _{r}^{\circ } = \left [ -5042.5 ]\left -[ -187.8] = \left ( -4854.7kJ \right )$

To calculate the difference: ΔH-ΔE=Δ(PV)

We use the ideal gas equation: P.V = n.R.T

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T

Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹

Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = 9910.288 J = 9.91 kJ (∵ 1 kJ = 1000J )

8 0

3 years ago

Which acid/base pair will give an equivalence point that cannot be predicted solely by a general knowledge of acid and base stre

sleet_krkn [62]

The acid/base pair that give equivalence point that Cannot be predicted by general knowledge is NaOH and HCI ONH.

An Acid is a substances that is corrosive in nature and turn blue lithmus paper to red which it react with base to produce salt and water.

Acid dissolve metals.

Base is a substance that turn red lihthmus paper to blue and react with acid to produce salt and water.

Therefore, The acid/base pair that give equivalence point that Cannot be predicted by general knowledge is NaOH and HCI ONH.

The question is incomplete as the options were not given. The options were gotten from another website.

Select the correct answer below:

ONaOH and HCI ONH,

HC ONH, and CH, COOH

NaOH and Christmas, COOH

Learn more about acid and base below.

brainly.com/question/2506771

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2 years ago

3. longitude is measured in degrees, hours, and seconds. a. true b. false

katen-ka-za [31]

Degrees true hours false seconds false

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3 years ago