Answer is: the missing pressure is 1088.66 mmHg.
Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
p₁/T₁ = p₂/T₂.
p₁ = 960 mmHg; pressure of the gas.
T₁ = 100°C + 273.15.
T₁ = 373.15 K; temperature of the gas.
T₂ = 150°C + 273.15.
T₂ = 423.15 K.
p₂ = p₁T₂/T₁.
p₂ = 960 mmHg · 423.15 K / 373.15 K.
p₂ = 1088.66 mmHg.
The products will be magnesium phosphate and potassium chloride. You then have to watch a solubility chart to see which one of these is not soluable. In this case it is magnesium phosphate.
The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O
Answer:
Ag+(aq) + Cl-(aq) —> AgCl(s)
Explanation:
2AgNO3(aq) + CaCl2(aq) —>2AgCl(s) + Ca(NO3)2(aq)
The balanced net ionic equation for the reaction above can be obtained as follow:
AgNO3(aq) and CaCl2(aq) will dissociate in solution as follow:
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
CaCl2(aq) —> Ca2+(aq) + 2Cl-(aq)
AgNO3(aq) + CaCl2(aq) –>
2Ag+(aq) + 2NO3-(aq) + Ca2+(aq) + 2Cl-(aq) —> 2AgCl(s) + Ca2+(aq) + 2NO3-(aq)
Cancel out the spectator ions i.e Ca2+(aq) and 2NO3- to obtain the net ionic equation.
2Ag+(aq) + 2Cl-(aq) —> 2AgCl(s)
Divide through by 2
Ag+(aq) + Cl-(aq) —> AgCl(s)
The, the net ionic equation is
Ag+(aq) + Cl-(aq) —> AgCl(s)
Answer:
4.6 × 10²³ molecules:
Step-by-step solution
You will need a balanced equation with masses, moles, and molar masses, so let's gather the information in one place:
M_r: 22.99
2Na + 2H₂O ⟶ 2NaOH + H₂
m/g: 35
1. Calculate the <em>moles of Na
</em>
Moles of Na = 35 g Na × (1 mol Na/22.99 g Na)
Moles of Na = 1.52 mol Na
2. Calculate the <em>moles of H₂
</em>
Moles of H₂ = 1.52 mol Na × (1 mol H₂/2 mol Na)
Moles of H₂= 0.761 mol H₂
3. Calculate the molecules of H₂
6.022 × 10²³ molecules H₂ = 1 mol H₂
Molecules of H₂ = 0.761 × (6.022 × 10²³
/1)
Molecules of H₂ = 4.6 × 10²³ molecules H₂
The reaction forms 4.6 × 10²³ molecules of H₂.
