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2 years ago

The accepted value is 29.35. Which correctly describes this student's experimental data?

Chemistry

1 answer:

kkurt [141]2 years ago

3 0

Answer:

- Both accurate and precise.

Explanation:

The experimental data is characterized as the data that is generated through a specific test or measurement with the purpose of analysis or evaluation. As per the question, the minuteness of the given data(29.27, 29.32, and 29. 27)most aptly describes that the experimental data collected by the student possesses 'accuracy and precision' apt and specific. It would yield the most worthy and reliable conclusion.

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Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this

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Answer: d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

$2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H=+1411.1kJ$

Reversing the reaction, changes the sign of $\Delta H$

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$

$\Delta H=-1411.1kJ$

On multiplying the reaction by $\frac{1}{2}$ , enthalpy gets half:

$0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)$ $\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol$

Thus the enthalpy change for the given reaction is -705.55kJ

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2 years ago

In an experiment, shinning which type of light on a strip of metal would be least likely to produce the photoelectric effect?

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C red light isn't an efficient source of light

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Answer:Ian is looking at cells using a microscope. He sees a nucleus and a large vacuole in the central area of a cell. What type of cell is he most likely looking at?

Explanation:

Ian is looking at cells using a microscope. He sees a nucleus and a large vacuole in the central area of a cell. What type of cell is he most likely looking at?

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2 years ago

Nitric oxide (NO) reacts readily with chlorine gas as follows.2 NO(g) + Cl2(g) equilibrium reaction arrow 2 NOCl(g)At 700. K the

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Answer:

For a: The mixture will need to produce more reactants to reach equilibrium.

For b: The mixture will need to produce more reactants to reach equilibrium.

For c: The mixture will need to produce more products to reach equilibrium.

Explanation:

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

The expression of $K_{p}$ for above equation follows:

$K_{p}=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$ .....(1)

We are given:

Value of $K_p$ = 0.26

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

For the given options:

For a:

We are given:

$p_{NOCl}=0.11atm\\p_{NO}=0.16atm\\p_{Cl_2}=0.30atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.11)^2}{(0.16)^2\times 0.30}=1.57$

As, $K_{p}$ ; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

For b:

We are given:

$p_{NOCl}=0.048atm\\p_{NO}=0.12atm\\p_{Cl_2}=0.10atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.048)^2}{(0.12)^2\times 0.10}=1.6$

As, $K_{p}$ ; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

For c:

We are given:

$p_{NOCl}=5.20\times 10^{-3}atm\\p_{NO}=0.15atm\\p_{Cl_2}=0.15atm$

Putting values in expression 1, we get:

$Q_p=\frac{(5.20\times 10^{-3})^2}{(0.15)^2\times 0.15}=0.008$

As, $K_{p}>Q_p$ ; the reaction is product favored.

Hence, the mixture will need to produce more products to reach equilibrium.

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olga nikolaevna [1]

Answer:

Anode:

3Mg(s) ----------> 3Mg2+(aq) + 6e

Cathode:

2Al3+(aq) +6e ---------> 2Al(s)

Explanation:

Anode:

3Mg(s) ----------> 3Mg2+(aq) + 6e

Cathode:

2Al3+(aq) +6e ---------> 2Al(s)

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