Answer: There are 
of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg
Explanation:
According to ideal gas equation:
P = pressure of gas = 895 mm Hg= 1.18 atm (760 mm Hg= 1 atm)
V = Volume of gas = 9.55 ml = 0.00955 L (1 L=1000ml)
n = number of moles = ?
R = gas constant =
T =temperature =
Thus there are of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg
Answer:
D) 65.7%
Explanation:
Based on the reaction:
2H2(g)+O2(g)⟶2H2O(l)
<em>2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.</em>
<em />
To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.
<em>Theoretical yield:</em>
Moles of 5.58g H₂:
5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂
As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:
2.768 moles H₂O ₓ (18.015g / mol) =
49.86g H₂O is theoretical yield
<em>Percent yield:</em>
Percent yield = Actual yield / Theoretical yield ₓ 100
32.8g H₂O / 49.86g ₓ 100 =
65.7% is percent yield of the reaction
<h3>D) 65.7%
</h3>
The right answer for the question that is being asked and shown above is that: "(2) the cathode in a voltaic cell and the anode in an electrolytic cell." At the status of electrode does oxidation occur in a voltaic cell and in an electrolytic cell is that the cathode in a voltaic cell and the anode in <span>an electrolytic cell</span>
Ok so first you need to figure out the energy of ONE photon with that wavelength. Using E=hc/lambda, you get E= 1.99 * 10^-20 J/photon. Now, how many photons do you need to add up to get to one kilojoule=1000 joules? 1000J / (1.99 * 10^-20 J/photon) = approximately 5 * 10^22 photons
hope this helps
