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kaheart [24]
3 years ago
8

Which of the following describes the thermal energy of particles in a substance?

Chemistry
1 answer:
VikaD [51]3 years ago
6 0
D I  hope is the answer
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Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O
pashok25 [27]

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S    = 10.8% => 10.8g/32g/mol = 0.34mol

%O   = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

3 0
3 years ago
The following information is given for benzene, C6H6, at 1atm: boiling point = 80.1 °C Hvap(80.1 °C) = 30.7 kJ/mol specific heat
user100 [1]

<u>Answer:</u> The heat required for the process is 4.24 kJ

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of benzene = 24.8 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

\text{Moles of benzene}=\frac{24.8g}{78.11g/mol}=0.318mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = ?

n = number of moles = 0.318 moles

\Delta H_{rxn} = enthalpy change of the reaction  = 30.7 kJ/mol

Putting values in above equation, we get:

30.7kJ/mol=\frac{q}{0.318mol}\\\\q=(30.7kJ/mol\times 0.318mol)=4.24kJ

Hence, the heat required for the process is 4.24 kJ

6 0
3 years ago
Choose the value for rotation to place triangle ABC on top of triangle DEF. How can you complete the transformation in only one
MariettaO [177]

Answer:90°

Explanation:

7 0
2 years ago
Read 2 more answers
Diamond cannot conduct electricity but graphite can,why​
zavuch27 [327]

Diamond does not conduct electricity because it has no delocalized electron to conduct electricity.

8 0
2 years ago
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×
Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0
3 years ago
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