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3 years ago

What is the volume of gas at the temperature of 100K

1 answer:

4 0

Answer:Sol. P base 1 = 10 kpa = 10 × 10^3 pa. P base 2 = 50 × 10^3 pa. v base 1 = 200 cc. v base 2 = 50 cc (i) Work done on the gas = ½ (10+50) ...

Explanation:

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AlF3 is almost insoluble in anhydrous HF but dissolves if KF is present.passage of BF3 through the resulting solution causes AlF

miskamm [114]

It's because the product formed with BF3 is more complex which able to decompose AlF3.

AlF3 doesn't dissolve in HF because of the fluorine. It's doesn't allow for coordination due to the hydrogen. However, it will dissolve in KF. If you look at the chemical reaction, it's able to form a salt.

3KF+AlF3−>3KF.AlF3

4 0

4 years ago

At what temperature,in degrees°C, would 12.6g of HCN occupy 13.3L if the pressure is 0.97 atm

Flura [38]

Answer:

63.5°c

Explanation:

almost postive this is right

5 0

4 years ago

Read 2 more answers

0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(

Studentka2010 [4]

Answer : The concentration of $Fe^{3+}$ at equilibrium is 0 M.

Solution : Given,

Concentration of $Fe^{3+}$ = 0.0200 M

Concentration of $C_2O_4^{2-}$ = 1.00 M

The given equilibrium reaction is,

$Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)$

Initially conc. 0.02 1.00 0

At eqm. (0.02-x) (1.00-3x) x

The expression of $K_c$ will be,

$K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}$

$1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}$

By solving the term, we get:

$x=0.02M$

Concentration of $Fe^{3+}$ at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of $Fe^{3+}$ at equilibrium is 0 M.

4 0

3 years ago

If a 2 kg object produces a 16 N force, what is its acceleration?

dolphi86 [110]

Answer:

acceleration (a)=8 m/s square.

7 0

3 years ago

The thiocyanate polyatomic ion, SCN-, is commanly called a pseudohalogen because it acts very much like halide ions. For example

harina [27]

Answer:

A) $\frac{1mol(SCN)_{2}}{2molNaSCN}$

B) 0.025 mol (SCN)₂

C) 2 mol (SCN)₂

D) $\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}$

E) 3.5504 mol H₂SO₄

Explanation:

2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄

A) A conversion factor could be $\frac{1mol(SCN)_{2}}{2molNaSCN}$ , as it has the units that we want to convert to in the numerator, and the units that we want to convert from in the denominator.

B) 0.05 mol NaSCN * $\frac{1mol(SCN)_{2}}{2molNaSCN}$ = 0.025 mol (SCN)₂

C) With 4 moles of NaSCN and 3 moles of MnSO₄, the reactant is NaSCN so we use that value to calculate the moles of product formed:

4 mol NaSCN * $\frac{1mol(SCN)_{2}}{2molNaSCN}$ = 2 mol (SCN)₂

D) $\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}$

E) 1.7752 mol MnSO₄ * $\frac{2molH_{2}SO_{4}}{1molMnSO_{4}}$ = 3.5504 mol H₂SO₄

5 0

3 years ago