Answer: 48.95g
Explanation:
no. of moles of Cl2 = 39/(2*35.5) = 0.55 mol
no. of moles of Al = 34/27 = 1.26 mol
hence, aluminium is in excess so we'll do calculation using no. of moles of Cl2 as it will be the only reactant to be used up completely. So,
no of moles of AlCl3 = 2/3 * (0.55) = 0.367 mol
hence amount of AlCl3 = 0.367 * (27+3*35.5) = 48.95g
You have to switch the elements based in if they're metals/nonmetals. so nonmetals switch with each other and metals switch with each other.
AB + CD ---> AC + BD
Answer:
The sample will be heated to 808.5 Kelvin
Explanation:
Step 1: Data given
Volume before heating = 2.00L
Temperature before heating = 35.0°C = 308 K
Volume after heating = 5.25 L
Pressure is constant
Step 2: Calculate temperature
V1 / T1 = V2 /T2
⇒ V1 = the initial volume = 2.00 L
⇒ T1 = the initial temperature = 308 K
⇒ V2 = the final volume = 5.25 L
⇒ T2 = The final temperature = TO BE DETERMINED
2.00L / 308.0 = 5.25L / T2
T2 = 5.25/(2.00/308.0)
T2 = 808.5 K
The sample will be heated to 808.5 Kelvin
The complete balanced chemical reaction is:
2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S
First let us calculate the number of moles of AgNO3.
moles AgNO3 = 0.315 M * 0.035 L
moles AgNO3 = 0.011025 mol
From the reaction, 1 mole of Na2S is needed for every 2
moles of AgNO3 hence:
moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2
mol AgNO3)
moles Na2S required = 5.5125 x 10^-3 mol
Therefore volume required is:
volume Na2S = 5.5125 x 10^-3 mol / 0.260 M
<span>volume Na2S = 0.0212 L = 21.2 mL</span>
