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Finger [1]
3 years ago
14

A certain amount of gas is trapped in a nonrigid container at atmospheric pressure. If the container is cooled and the gas in th

e container remains at atmospheric pressure, what will happen? A. The number of moles of gas will decrease. B. The number of moles of gas will increase. C. The temperature of the gas will increase. D. The volume of the container will decrease. E. The volume of the container will increase.
Chemistry
1 answer:
Akimi4 [234]3 years ago
8 0

Answer:

D. The volume of the container will decrease.

Explanation:

  • We apply here, the general gas law of ideal gases:

<em>PV = nRT,</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas.

  • We are given that P is constant.

Also, n is constant as there is no added external gases.

T is decreased as the container is cooled.

So, the volume will change as the container is nonrigid.

<em>V α T.</em>

The volume is directly proportional to the temperature,

<em>Thus, as the container is cooled, the T will decrease, and the volume of the container will decrease.</em>

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Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
Maru [420]

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and 1s^22s^22p^6

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1

This element will easily loose 1 electron and form Rb^+ ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

1s^22s^22p^3

This element will easily gain 3 electrons and form N^{3-} ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

1s^22s^22p^6

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^5

This element will easily gain 1 electron and form Br^- ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

4 0
3 years ago
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
If 30.943 g of a liquid occupy a space of 35.0 ml. What is the density of the liquid in g/cm3?
timofeeve [1]
Mass / volume = density
30.943g / 35ml = 0.88408571g/ml 

7 0
3 years ago
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Benzene would react_____________.
Vlada [557]

Answer: Benzene is less reactive than methylbenzoate and more reactive than Nitrobenzene

Explanation:

This is because the methyl group on the benzene ring is an electron donating group leading to the activation of the ring and subsequently leading to more canonical resonance structure at the intermediate stage of the reaction enhancing the faster reactivity

However for the Nitrobenzene the nitro group is an electron withdrawing group leading to a slower activation and less resonance canonical structure at the reaction intermediate leading to a slower reaction than the reaction of benzene without the nitro group

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3 years ago
When the following equation is balanced, what is the coefficient for HBr?
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