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3 years ago

PLEZ HELP QUICKLY!!!! ILL MARK BRAINLIEST

Chemistry

2 answers:

stellarik [79]3 years ago

5 0

It is C.ti and o and you don’t have to make me brainlest

Readme [11.4K]3 years ago

3 0

Option B, P and Cl

They form covalent bond

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Which of these statements is false?

AVprozaik [17]

Answer:

The answer is b

Explanation:

Hope it helps

8 0

3 years ago

Read 2 more answers

g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the

weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0

3 years ago

A system gains 652 kJ of heat, resulting in a change in internal energy of the system equal to +241 kJ. How much work is done?

levacccp [35]

Answer:

-411 kj

Explanation:

We solve by using this formula

∆U = ∆Q + ∆W

This formula is the first law of thermodynamics

Change in internal energy U = +241

Heat gained by system Q = 652

Putting the value into the equation

+241 = 652 + W

Workdone = 241 - 652

Workdone = -411 kj

Since work done is negative it means that work was done by the system

3 0

3 years ago

1.72 moles of NOCI were placed in a 2.50 L reaction chamber

elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

2NOCI(g) = 2NO(g) + Cl2(g).

I 1.72

C 0.56 0.56 0.28

E 1.16 0.56 0.28

Molarity at equilibrium :

NOCl :

$\tt \dfrac{1.16}{2.5}=0.464$

NO :

$\tt \dfrac{0.56}{2.5}=0.224$

Cl2 :

$\tt \dfrac{0.28}{2.5}=0.112$

$\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026$

4 0

3 years ago

PLEASEEEEEEEEEEEE HELPPPPPPPP I BEGGGGG FOR HELPPPPP

Elza [17]

Answer: There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles$

1 mole of $H_2O$ contains = $6.023\times 10^{23}$ molecules

Thus 3.5 moles of $H_2O$ contains = $\frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23}$ molecules.

There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

3 0

3 years ago