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yuradex [85]
3 years ago
12

Why does warm air raise and cold air sink

Chemistry
2 answers:
Leni [432]3 years ago
6 0
Warm air is lighter than cold air

Did I help? Thank me by clicking Best Answer!
ki77a [65]3 years ago
3 0
Warm is less heaver then cold air so warm air rise and cold air sinks
  
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MAGNESIUM (Mg)<br> +FLUORINE (F) =
bazaltina [42]

Answer:

MgF2 magnesium fluoride

4 0
3 years ago
What is the percent yield of ferrous sulfide if the actual yield is 220.0g and the theoretic yield is 275.6?
Amanda [17]
ANSWER yield = 79.83%
3 0
3 years ago
in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
The mass of an atom may be found by adding the
Darina [25.2K]

Answer:

protons and neutrons

Explanation:

5 0
2 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
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