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3 years ago

Question 9 (1 point)

Chemistry

1 answer:

Juliette [100K]3 years ago

7 0

Noble gases

Explanation:

Electronic configuration 1s² 2s² 2p⁶

The element belongs to the group of the noble gases.

The noble gases have complete outer shell configuration of their atoms.
we can infer that the configuration above is for an element in the p-block because the last sub-level filled is the p-orbital.
The elements therefore belongs to the p-block
The block is from group 111A to O
Only the halogens and noble gases fits this picture from the option.
The outer most p-subshell have three orbitals requiring 6 electrons to fill them up.
This makes a complete and stable configuration.
The highest energy level of 2 is also made up of 8 electrons, an octet.
This is why we can conclude that they are noble gases.

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If you observe something and want to determine why it is happening, what steps should you

Sloan [31]

Answer:

Make an observation.

Ask a question.

Form a hypothesis, or testable explanation.

Make a prediction based on the hypothesis.

Test the prediction.

Iterate: use the results to make new hypotheses or predictions.

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3 years ago

The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2

notsponge [240]

Answer : The unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law expression for the reaction:

$\text{Rate}=k[N_2]^a[H_2]^b$

where,

a = order with respect to $N_2$

b = order with respect to $H_2$

Expression for rate law for first observation:

$0.0021=k(0.10)^a(0.10)^b$ ....(1)

Expression for rate law for second observation:

$0.0084=k(0.10)^a(0.20)^b$ ....(2)

Expression for rate law for third observation:

$0.0672=k(0.20)^a(0.40)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2$

Dividing 3 by 1 and also put value of b, we get:

$\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[N_2]^a[H_2]^b$

$\text{Rate}=k[N_2]^1[H_2]^2$

Now, calculating the value of 'k' by using any expression.

$0.0021=k(0.10)^1(0.10)^2$

$k=2.1M^{-2}min^{-1}$

The value of the rate constant 'k' for this reaction is $2.1M^{-2}min^{-1}$

That means, the unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

7 0

3 years ago

When a chemist collects hydrogen gas over water, she ends up with a mixture of hydrogen and water vapor in her collecting bottle

elixir [45]

<u>Answer:</u> The partial pressure of hydrogen is 93.9 kPa.

<u>Explanation:</u>

To calculate the partial pressure of hydrogen, we will follow Dalton's Law.

This law states that the total pressure of a mixture of gases is equal to the sum of the individual pressures exerted by the constituent gases.

Mathematically,

$p_{total}=p_A+p_B$

According to the question,

$p_{total}=p_{H_2O}+p_{H_2}$

We are given:

$p_{total}=97.1 kPa$

$p_{H_2O}=3.2kPa$

$p_{H_2}= ?kPa$

Putting values in above equation, we get:

$97.1kPa=3.2kPa+p_{H_2}$

$p_{H_2}= 93.9kPa$

Hence, the partial pressure of hydrogen is 93.9 kPa.

5 0

3 years ago

Read 2 more answers

how many monochloro substitution products are produced when the alkanes below are chlorinated? consider constitutional isomers o

Airida [17]

When the alkanes below are chlorinated they produce 3 monochloro substitution products. These isomers have the same chemical formula, but their atoms arrangements are different.

What are constitutional versus isomeric isomers?

Although structural (constitutional) isomers share the same chemical formula but their atoms are bonded in a different way. Stereoisomers have the same atomic configurations and chemical formulae. The groups' spatial orientation is the only thing which distinguishes the molecules distinguishes them from one another.

What are instances of stereoisomers?

According to the general definition of stereoisomers, stereoisomers are isomers with the same composition (i.e., the same parts), but different orientations in space. Stereoisomers are of two different types which are enantiomers and diastereomers.

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1 year ago

Convert each quantity to the indicated units. a. 3.01g to cg. b. 6200m to km

miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if 1 g is equal to 100 cg

then 3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if 1 km is equal to 1000 m

then Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

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3 years ago