We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
You can learn more about solutions here: brainly.com/question/2412491
The International System of Units is the current international standard metric system and is also the system most widely used around the world.
Answer:
2666.7 hours
Explanation:
The key to solve this problem is that we are given the propane gas consumed in one hour by giving us the information of the volume consumed at 1 atm, 298 K (25 +273). Using the gas law we can calculate the rate of consumption of propane per hour, and from here we can calculate its mass and converting it to gallons and finally diving the 400 gallos by this number.
PV = nRT ∴ n = PV/RT
n = 1 atm x 165 L/ (0.08206 Latm/kmol x 298 K ) = 6.75 mol propane
Mass propane :
6.75 mol x 44 g/mol = 296.88 g
convert this to Kg:
296.88 g/ 1000 g/Kg = 0.30 Kg
calculate the volume in liters this represents by dividing by the density:
0.30 Kg / 0.5077 Kg/L = 0.59 L
changing this to gallons
0.59 L x 1 gallon/3.785 L = 0.15 gallon
and finally calculate how many hours the 400 gallons propane tank will deliver
400 gallon/ 0.15 gallon/hr = 2666.7 hr
The percentage yield is 72.8 %.
<em>Step 1</em>. Calculate the <em>mass of Br₂</em>
Mass of Br₂ = 20.0 mL Br₂ × (3.10 g Br₂/1 mL Br₂) = 62.00 g Br₂
<em>Step 2</em>. Calculate the <em>theoretical yield</em>
M_r: 159.81 266.69
2Al + 3Br₂ → 2AlBr₃
Moles of Br₂ = 62.00 g Br₂ × (1 mol Br₂/(159.81 g Br₂) = 0.3880 mol Br₂
Moles of AlBr₃ = 0.3880 mol Br₂ × (2 mol AlBr₃/(3 mol Br₂) = 0.2586 mol AlBr₃
Theor. yield of AlBr₃ = 0.2586 mol AlBr₃ × 266.99 g AlBr₃)/(1 mol AlBr₃)
= 69.05 g AlCl₃
<em>Step 3</em>. Calculate the <em>percentage yield
</em>
% yield = (actual yield/theoretical yield) × 100 % = (50.3 g/69.05 g) × 100 %
= 72.8 %
Answer:
3.5 g
Explanation:
The density of water is 1 g/mL. The mass (m) corresponding to 20.0 mL is 20.0 g.
We can calculate the heat (Q) required to raise the temperature of 20.0 mL of water 1 °C (ΔT).
Q = c × m × ΔT = 1 cal/g.°C × 20.0 g × 1 °C = 20 cal
where,
c is the specific heat capacity of water
There are 160 calories in 28 g of Cheetos. The mass that releases 20 cal is:
20 cal × (28 g/160 cal) = 3.5 g
