CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
Answer:
PH= 6.767 (answer is the A option)
Explanation:
first we need to correct the value in Kw at this temperature is 2.92*10^-14
so, in this case we have that:
Kw=2.92*10^-14 M²
[ H3O^+] [ H3O^+]
![[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%5BOH%5E%7B-%7D%20%20%5D%20%3D%20Kw%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D%20%20%20%5C%5C%5C%5C)
at 40ºC
![[H_{3}O^{+} ] = [OH^{-} ]](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%5BOH%5E%7B-%7D%20%20%5D)
![[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%5E%7B2%7D%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D)
![[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%282.92%2A10%5E%7B-14%7D%29%5E%7B1%2F2%7D%20%3D%201.71%2A10%5E%7B-7%7D%20M)
![PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767](https://tex.z-dn.net/?f=PH%3D%20-log10%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20-log10%281.71%2A10%5E%7B-7%7D%20%29%20%3D%206.767)
<h3>
Answer:</h3>
83.33 seconds.
<h3>
Explanation:</h3>
<u>We are given;</u>
- Take off velocity as 300 km/hr
- Acceleration as 1 m/s²
We are required to calculate the take off time of the airplane.
<h3>Step 1: Convert velocity from km/hr to m/s </h3>
We are going to use the conversion factor.
The conversion factor is 3.6 km/hr per m/s
Therefore;
Velocity = 300 km/hr ÷ 3.6 km/hr per m/s
= 83.33 m/s
<h3>Step 2: Calculate the take off time</h3>
We know that;
v = u + at
where, u is the initial velocity, v the final velocity, a the acceleration and t is time.
But, initial velocity is Zero
Therefore;
83.33 m/s = 1 m/s² × t
Thus;
time = 83.33 m/s ÷ 1 m/s²
= 83.33 seconds
Therefore, the take off time is 83.33 seconds.
Answer:
it depends on the subject but i can see what i can do
Explanation:
Answer:
Бардык белгилүү авиация мыйзамдарына ылайык, аары учуу мүмкүнчүлүгү жок. Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.
Explanation:
Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.