To solve this problem it is fundamentally, just look for the volume of the gas and convert it to cm3. At STP 1 mole = 22.4 liters. 8.00 moles x 22.4 liters/mole = 179.2 liters = 179,200 cm^3 Then. get the cube root of 179,200 cm^3. This would be equal to 56.38 cm and thus would be the length of the edge of this cube.
Answer:
2.038 seconds.
Explanation:
So, in the question above we are given the following parameters in order to solve this question. We are given a rate constant of 0.500 s^-, initial concentration= 0.860 M and final concentration= 0.310 M,the time,t =??.
Assuming that the equation for the first order of reaction is given below,that is;
A ---------------------------------> products.
Recall the formula below;
B= B° e^-kt.
Therefore, e^-kt = B/B°.
-kt = ln B/B°.
kt= ln B°/B.
Where B° and B are the amount of the initial concentration and the amount of the concentration remaining, k is the rate constant and t = time taken for the concentration to decrease.
So, we have; time taken,t = ln( 0.860/.310)/0.500.
==> ln 2.77/0.500.
==> time taken,t =2.038 seconds.
Answer: bro pic is not cleyer
Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm
Answer:
eight
Explanation:
Hence, from the above given configuration it is clear that the number of valence electrons in the Argon atom is eight.
