Answer:
i dont speak your language im so sorry
Explanation:
Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is 
Explanation:
The relation of density and molar mass is:
where
d = density = 3.27 g/ L
P = pressure of the gas = 773 torr = 1.02 atm (760 torr = 1atm)
M = molar mass of the gas = ?
T = temperature of the gas = 
R = gas constant = 
The relation of depression in freezing point with molality:
= depression in freezing point = 
= 
= freezing point constant = 5.1
m = molality = 
Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is
The chemical reaction for the ammonium ion in water is:
NH4+(aq) + H2O(l) ----> H3O+ (aq) + NH3 (aq) Ka = <span>5.60x10^-10
The ammonium ion NH4+ in aqueous form reacts with liquid water forming hydronium ion (aqueous) and aqueous ammonia, with a Ka of 5.60x10^-10.</span>
<h3>
Answer:</h3>
Initial temperature is 243.59°C
<h3>
Explanation:</h3>
The quantity of heat is calculated by multiplying the mass of a substance by its specific heat capacity and change in temperature.
That is; Q = m×c×ΔT
In this case;
Quantity of heat = 560 J
Mass of the Sample of Zinc = 10 g
Final temperature = 100°C
We are required to determine the initial temperature;
This can be done by replacing the known variables in the formula of finding quantity of heat,
Specific heat capacity, c, of Zinc = 0.39 J/g.°C
Therefore,
560 J = 10 g × 0.39 J/g°C × ΔT
ΔT = 560 J ÷ (3.9 J/°C)
= 143.59°C
But, since the sample of Zinc lost heat then the temperature change will have a negative value.
ΔT = -143.59°C
Then,
ΔT = T(final) - T(initial)
Therefore,
T(initial) = T(final) - ΔT
= 100°C - (-143.59°C)
= 243.59°C
Hence, the initial temperature of zinc sample is 243.59°C
Answer:
0.800 mol
Explanation:
We have the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with moles of the compounds involved.
Step 1. <em>Gather all the information</em> in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 4.00 4.00
===============
Step 2. Identify the <em>limiting reactant
</em>
Calculate the <em>moles of CO₂</em> we can obtain from each reactant.
<em>From C₃H₈:</em>
The molar ratio of CO₂: C₃H₈ is 3:1
Moles of CO₂ = 4.00 × 3/1
Moles of CO₂ = 12.0 mol CO₂
<em>From O₂</em>:
The molar ratio of CO₂: O₂ is 3:5.
Moles of CO₂ = 4.00 × ⅗
Moles of CO₂ = 2.40 mol CO₂
O₂ is the limiting reactant because it gives the smaller amount of CO₂.
==============
Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.
The molar ratio of C₃H₈:O₂ is 1:5.
Moles of C₃H₈ = 4.00 × ⅕
Moles of C₃H₈ = 0.800 mol C₃H₈
