Answer:
See explanation
Explanation:
The equation of the reaction is;
C3H8 + 5O2 ----> 3CO2 + 4H2O
Number of moles of C3H8 = 132.33g/44g/mol = 3 moles
1 mole of C3H8 yields 3 moles of CO2
3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2
Number of moles of oxygen = 384.00 g/32 g/mol = 12 moles
5 moles of oxygen yields 3 moles of CO2
12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2
Hence C3H8 is the limiting reactant.
Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2
1 moles of C3H8 yields 4 moles of water
3 moles of C3H8 yields 3 × 4/1 = 12 moles of water
Mass of water = 12 moles of water × 18 g/mol = 216 g of water
b) Actual yield = 269.34 g
Theoretical yield = 396 g
% yield = actual yield/theoretical yield × 100/1
% yield = 269.34 g /396 g × 100
% yield = 68%
The equilibrium membrane potential is 41.9 mV.
To calculate the membrane potential, we use the <em>Nernst Equation</em>:
<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}
where
• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions
• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]
• <em>T</em> = the Kelvin temperature
• <em>z</em> = the charge on the ion (+1)
• <em>F </em>= the Faraday constant [96 485 C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]
• [Na]_o = the concentration of Na^(+) outside the cell
• [Na]_i = the concentration of Na^(+) inside the cell
∴ <em>V</em>_Na =
[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV
A compound contains two or more elements chemically combined
Stir the water continuously, this is the only logical answer. Adding powdered sugar, decreasing the volume, increasing the amount of suga, cooling, all don't make the sugar dissolve quicker.
Answer:
Tungsten is used for this experiment
Explanation:
This is a Thermal - equilibrium situation. we can use the equation.
Loss of Heat of the Metal = Gain of Heat by the Water

Q = mΔT
Q = heat
m = mass
ΔT = T₂ - T₁
T₂ = final temperature
T₁ = Initial temperature
Cp = Specific heat capacity
<u>Metal</u>
m = 83.8 g
T₂ = 50⁰C
T₁ = 600⁰C
Cp = 
<u>Water</u>
m = 75 g
T₂ = 50⁰C
T₁ = 30⁰C
Cp = 4.184 j.g⁻¹.⁰c⁻¹

⇒ - 83.8 x
x (50 - 600) = 75 x 4.184 x (50 - 30)
⇒
=
j.g⁻¹.⁰c⁻¹
We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹
So metal Tungsten used in this experiment