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3 years ago

Consider the following: If the Earth's Crust are constantly moving, what

Chemistry

1 answer:

Vesnalui [34]3 years ago

5 0

Answer:

the shape of the continents will continue to change, without even accounting for rising sea levels

Explanation:

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Basically everything is in the picture, and i dont really understand please help.

zubka84 [21]

This is what I found

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3 years ago

¡cual es el % masa/volumen de una mezcla que posee 40 gramos de colorante en 5 litros de mezcla?

Firlakuza [10]

Answer:

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Explanation:

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3 years ago

Suppose of nickel(II) chloride is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of chl

stich3 [128]

Answer: Molarity of chloride anion = 0.32 M

<em>Note: the question is missing some values. The full question is given below;</em>

<em>Suppose 7.26 g of nickel(II) chloride is dissolved in 350 mL of a 0.50 M aqueous solution of potassium carbonate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.</em>

Explanation:

Molarity or molar concentration is the number of moles (mol) of component per volume (liters) concentration of solution in mol/L or M

The mass of nickel (II) chloride is 7.26 g.

The volume of potassium carbonate is 350 mL = 0.35 L

The molarity of potassium carbonate solution is 0.50 M

The reaction of nickel (II) chloride and potassium carbonate is given below.

NiCl₂(aq) + KCO₃(aq) --------> KCl(aq) +NiCO₃(s)

The dissociation of nickel (II) chloride is given below.

NiCl₂ -----> Ni²⁺ + 2Cl⁻

The molar mass of nickel (II) chloride is 129.6 g/mol

The moles of nickel (II) chloride can be calculated by the formula given below;

No of moles = mass(g) / molar mass (g/mol)

No of moles = 7.26 / 129.6 = 0.056 moles

Therefore, molarity of NiCl₂ = 0.056 moles/ 0.35 L = 0.16 M

The molarity of 1 mole nickel (ii) chloride is 0.16 m and according to dissociation of nickel (II) chloride, 1 mole of nickel (II) chloride gives 2 moles of chloride anion.

Therefore, the molarity of chloride anion = 0.16 * 2 = 0.32 M

3 0

3 years ago

PLEASE HELP ITS DUE TODAY!

Yuri [45]

Answer:

At one atmosphere and twenty-five degrees Celsius, could you turn it into a liquid by cooling it down? Um, and the key here is that the triple point eyes that minus fifty six point six degrees Celsius and it's at five point eleven ATMs. So at one atmospheric pressure, there's no way that you're ever going to reach the liquid days. So the first part of this question is the answer The answer to the first part of a question is no. How could you instead make the liquid at twenty-five degrees Celsius? Well, the critical point is at thirty-one point one degrees Celsius. So you know, if you're twenty-five, if you increase the pressure instead, you will briefly by it, be able to form a liquid. And if you continue Teo, you know, increase the pressure eventually form a salad, so increasing the pressure is the second part. If you increase the pressure of co two thirty-seven degrees Celsius, will you ever liquefy? No. Because then, if you're above thirty-one point one degrees Celsius in temperature. You'LL never be able to actually form the liquid. Instead, you'LL only is able Teo obtain supercritical co too, which is really cool thing. You know, they used supercritical sio tu tio decaffeinated coffee without, you know, adding a solvent that you'LL be able to taste, which is really cool. But no, you can't liquefy so two above thirty-one degrees Celsius or below five-point eleven atmospheric pressures anyway, that's how I answer this question. Hope this helped :)

5 0

2 years ago

The salt formed by the reaction of the weak acid hydrocyanic acid, HCN, with the strong base potassium hydroxide is

iogann1982 [59]

Answer:

2.28 × 10^-3 mol/L

Explanation:

The equation for the equilibrium is

CN^- + H2O ⇌ HCN + OH^-

Ka = 4.9 × 10^-10

KaKb = Kw

4.9 × 10^-10 Kb = 1.00 × 10^-14

Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5

Now, we can set up an ICE table

CN^- + H2O ⇌ HCN + OH^-

I/(mol/L) 0.255 0 0

C/(mol/L) -x +x +x

E/(mol/L) 0.255 - x x x

Ka = x^2/(0.255 - x) = 2.05 × 10^-5

Check for negligibility

0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6

x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3

[OH^-] = x mol/L = 2.28 × 10^-3 mol/L

5 0

3 years ago