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atroni [7]
3 years ago
9

How many single covalent bonds can halogens form?

Chemistry
1 answer:
Simora [160]3 years ago
4 0
One single covalent bond, hope this helps!
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11.<br> The electron configuration for phosphorous is [Ar]3s23p4.<br> TRUE<br> FALSE
user100 [1]

Answer:

False

Explanation:

Phosphorus is number 15 on the periodic table, so its electronic configuration is:

{1s}^{2} {2s}^{2} {2p}^{6} {3s}^{2} {3p}^{3}

5 0
2 years ago
Which of these 3 molecules is most polar? And which is the least polar? H2O, H2S, or SO2? Explain.
stepan [7]

Answer:so2

Explanation:

5 0
3 years ago
NaBr + CaF2 → NaF + CaBr2 What coefficients are needed to balance the chemical equation? A) 1,1,1,1 B) 1,2,1,2 C) 1,2,2,1 D) 2,1
elena-s [515]
D.
2NaBr + CaF2 --> 2NaF + CaBr2 gives you:

2Na                        2Na
2Br                         2F
1Ca                         1Ca
2F                           2Br

This is balanced.
7 0
3 years ago
Read 2 more answers
A mixture of He , Ar , and Xe has a total pressure of 3.00 atm . The partial pressure of He is 0.200 atm , and the partial press
yaroslaw [1]

The partial pressure of Xe is 2.60atm

The sum of the partial pressures is equal to the total pressure. Ptotal = P1 + P2 + P3... .To calculate the partial pressure of one of the components of a mixture, subtract the sum of known partial pressures of the other components from the total pressure.

Given the values of

PHe=0.20 atm

PAr=0.20 atm

Ptotal=3.00 atm

We need to find  PXe

Solution: Determine the sum of the partial pressures of helium and argon, then subtract the sum from the total pressure.

PHe+PAr =0.20 atm + 0.20=0.400 atm

PXe=3.00atm−0.400=2.60 atm

Each gas that makes up a mixture of gases has a partial pressure, which is the notional pressure of that gas as if it alone filled the original combination's complete volume at the same temperature.

According to Dalton's Law, a perfect gas mixture's total pressure equals the sum of its constituent gases' individual partial pressures.

The thermodynamic activity of a gas's molecules is gauged by its partial pressure.

Gases react, disperse, and dissolve based on their partial pressures rather than the concentrations they have in liquids or other gas combinations.

This general characteristic of gases holds true in biological chemical interactions involving gases.

Hence the partial pressure of Xe is 2.60atm

Learn more about partial pressure here

brainly.com/question/19813237

#SPJ4

5 0
2 years ago
Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
3 years ago
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