Answer: 5.66 dm3
Explanation:
Given that:
Volume of neon gas = ?
Temperature T = 35°C
Convert Celsius to Kelvin
(35°C + 273 = 308K)
Pressure P = 0.37 atm
Number of moles N = 0.83 moles
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.37atm x V = 0.83 moles x 0.0082 atm dm3 K-1 mol-1 x 308K
0.37 atm x V = 2.096 atm dm3
V = (2.096 atm dm3 / 0.37atm)
V = 5.66 dm3
Thus, the volume of the neon gas is 5.66 dm3
Answer:
120g
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction.
Sn + 2HF —> SnF2 + H2
Step 2:
Determination of the number of mole HF needed to react with 3 moles of Sn.
From the balanced equation above,
1 mole of Sn and reacted with 2 moles of HF.
Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.
Step 3:
Conversion of 6 moles of HF to grams.
Number of mole HF = 6 moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn.
Answer:
20ppm
Explanation:
parts per million are defined as the mass of solute in mg (In this case, mass of DDT) per kg of sample.
To solve this question we must find the mass of DDT in mg and the mass of sample in kg:
<em>Mass DDT:</em>
0.10g * (1000mg / 1g) = 100mg
<em>Mass sample:</em>
5000g * (1kg / 1000g) = 5kg
Parts per Million:
100mg / 5kg =
<h3>20ppm</h3>
Answer:
K8S4O16 or K8(SO4)4 depending on if the SO4 is supposed to represent sulfate or not
Explanation:
Find the molar mass of K2SO4 first:
2K + S + 4O ≈ 174 g/mol
Divide the goal molar mass of 696 by the molar mass of the empirical formula:
696 / 174 = 4
This means you need to multiply everything in the empirical formula by 4:
K2SO4 --> K8S4O16 or K8(SO4)4 depending on if the SO4 is for sulfate or not
Answer:
A. London dispersion
Explanation:
London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.
