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Katyanochek1 [597]
4 years ago
11

One of Mercury’s _________ properties is that it is a liquid at room temperature

Chemistry
1 answer:
larisa [96]4 years ago
7 0
The answer should be physical<span />
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Explain Boyle's pressure-volume relationship in terms of the kinetic-<br>molecular theory ​
Troyanec [42]

Answer:

The pressure and the volume are inversely related in Boyle's Law. In Kinetic theory pressure is created by the collision of particles. If the volume is greater the number of collisions and pressure will be less. So volume and pressure are inversely related if the temperature and total kinetic energy is kept constant.

Explanation:

hope this to help you

8 0
3 years ago
Read 2 more answers
Chlorine–35 has 17 protons. How many protons and neutrons does the isotope chlorine–36 have? 19 protons and 17 neutrons 17 proto
pochemuha

Answer:

17 protons

19 neutrons

Explanation:

Chlorine will always have the same amount of protons, and that would be 17 protons.

The atomic mass will change according to how many neutrons are present.

Cl - 35 is comprised of 17 protons and 18 neutrons.

We want to find Cl - 36:

We simply add 1 neutron. 18 + 1 = 19 neutrons.

8 0
4 years ago
Read 2 more answers
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
We kept adding pbi2 into water until no more will dissolve. what is the concentration of lead (ii) iodide in the solution
blsea [12.9K]

PbI(ii) ionization in the solution of PBI(ii) into water is:

<span>PbI</span>₂(solution) <==> Pb₂⁺ + 2I⁻

If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.

Therefore,

<span>Ksp=<span>[Pb</span></span>²⁺][I-]²

Plugging the values:

1.4×10⁻⁸ = x ⋅ (2x)²

1.4×10⁻⁸ = 4x³

x³ = {1.4×10⁻⁸}÷4

x³ = 0.35 x 10⁻⁸

or 

x³ = 3.5 x 10⁻⁹

x = 1.51 x 10⁻³

Hence,

Concentration of iodide ions in the solution:

2x = 3.02 x 10⁻³ 

6 0
3 years ago
Calculate the molality of a solution prepared by dissolving 25.0 g of sodium chloride (NaCl) in 625 g of water
umka2103 [35]

Answer: m= 0.69 m or mol/kg

Explanation: Molality is expressed as moles per unit kilograms of solvent or m= n / kg

First convert 25.0 g NaCl to moles

25.0 g NaCl x 1 mole NaCl / 58 g NaCl

= 0.43 moles NaCl

Next convert 625 g H2O to kilograms

625 g H2O x 1 kg / 1000 g H2O

= 0.625 kg H2O

Substitute the values in the formula

m = n / kg

   = 0.43 mole NaCl / 0.625 kg

   = 0.69 m or mole / kg

6 0
3 years ago
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