Which property of potential energy distinguishes it from kinetic energy

Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2Al(s)+3H2SO4(

ryzh [129]

Answer:

83.9g of sulfuric acid is the minimum mass you would need

1.73g of hydrogen would be produced

Explanation:

Based on the reaction:

2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

2 moles of solid aluminium react with 3 moles of sulfuric acid. Also, two moles of Al produce 3 moles of hydrogen gas.

15.4g of Al are:

15.4g Al × (1mol / 26.98g) = 0.571 moles of Al.

Moles of sulfuric acid:

0.571 moles Al × (3 mol H₂SO₄ / 2 mol Al) = 0.8565 moles H₂SO₄

In grams:

0.8565 moles H₂SO₄ × (98g / 1mol) = <em>83.9g of sulfuric acid is the minimum mass you would need</em>

In the same way, moles of hydrogen produced are:

0.571 moles Al × (3 mol H₂ / 2 mol Al) = 0.8565 moles H₂

In grams:

0.8565 moles H₂ × (2.015g / 1mol) = <em>1.73g of hydrogen would be produced</em>

3 0

3 years ago

Why can't we extract sodium from aqueous solution of sodium chloride

Mazyrski [523]

Answer:

As it is an aqueous solution the sodium is dissolved in the water, if you want to extract it use evaporation

Explanation:

3 0

2 years ago

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(c) $Ksp=2.27x10^{-12}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

(C) $NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

3 years ago

How much energy is required to heat 250 grams of water from 20.0°C to 80.0°C? The heat capacity of water is 4.186 J/g°C

tatyana61 [14]

Answer:

62,790J

Explanation:

Q=mcT

= 250g*4.186J/gc*80-20

=250*4.186*60=62,790J

7 0

3 years ago

Who can teach me chemical counting? please...

kupik [55]

Explanation:

sorry hindi ko alam eii!!!!

4 0

2 years ago