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3 years ago

What is the general trend for electron affinity values going across a period? The

Chemistry

2 answers:

KIM [24]3 years ago

8 0

Answer:

a.the values become more negative

Explanation:

k0ka [10]3 years ago

5 0

Answer: The values become more negative

Explanation: I just took the quiz and got it correct :)

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The metal rhenium becomes superconducting at temperatures below 1.4 K. Calculate the temperature at which rhenium becomes superc

kotegsom [21]

Answer:

The temperature at which rhenium becomes superconducting is -271.6 ºC.

Explanation:

We can convert the temperature given in Kelvin, which is an absolute scale, into degrees Celsius. which is a relative scale, by a simple algebraic translation, that is:

$T_{C} = T_{K} - 273$ (1)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{K}$ - Temperature, measured in Kelvins.

If we know that $T_{K} = 1.4\,K$ , then the temperature given in degrees Celsius is:

$T_{C} = 1.4\,K-273$

$T_{C} = -271.6\,^{\circ}C$

The temperature at which rhenium becomes superconducting is -271.6 ºC.

8 0

3 years ago

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

Answer: The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago

What location around the world are volcanoes consistently found

Nata [24]

The location that volcanoes are consistently found in the world is near the Ring of Fire so the answer would basically be "Continent-ocean boundaries".

4 0

4 years ago

Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad

ki77a [65]

Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O

20 mmol 36 mmol 20 mmol

16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO → Na⁺ + HCOO⁻

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So: [OH⁻] = 16 mmol / 400 mL → 0.04 M

- log [OH⁻] = pOH → 1.39

pH = 14 - pOH → 12.61

6 0

3 years ago

ID: 16. A mineral is always a(n) because it has a definite volume and shape.

Annette [7]

Answer:

Solid

Explanation:

3 0

3 years ago

Read 2 more answers