We are given with a compound, Methane (CH4), with a molar
mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in
g. We need to solve first the molecular weight of Methane, that is
C=12 g/mol
H=1g/mol
CH4= 12 g/mol +1(4) g/mol = 16 g/mol
With 0.893 mol sample, its corresponding mass is
g CH4= 0.893 mol x 16g/mol =14.288 g
Therefore, the mass of methane is 14.288 g
Answer:
d. One single bond and two double bonds.
Explanation:
The octate rule is a chemical rule in which the atoms prefer to have eight electrons in the valence shell. Where a single bond provide two electrons and a double bond provide 4 electrons. Thus:
a. Two double bonds
. Two double bonds provide 8 electrons. Octate rule <em>is not </em>violated
b. Three single bonds and one pair of electrons
. Three single bonds provide 6 electrons and one pair of electrons provide two electrons. Thus, you have eight electrons and octate rule <em>is not</em> violated
c. Two single bonds and one double bond
. Two single bonds provide four electrons and one double bond 4. Thus, you have eight electrons and octate rule <em>is not </em>violated.
d. One single bond and two double bonds. One single bond provides two electrons and two double bonds 8. Thus, you have 10 electrons and <em>octate rule is violated.</em>
e. Four single bonds. Four single bonds provide 8 electrons. Octate rule<em> is not </em>violated.
I hope it helps!
Answer:
3.24 × 10^5 J/mol
Explanation:
The activation energy of this reaction can be calculated using the equation:
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
Where; Ea = the activation energy (J/mol)
R = the ideal gas constant = 8.3145 J/Kmol
T1 and T2 = absolute temperatures (K)
k1 and k2 = the reaction rate constants at respective temperature
First, we need to convert the temperatures in °C to K
T(K) = T(°C) + 273.15
T1 = 325°C + 273.15
T1 = 598.15K
T2 = 407°C + 273.15
T2 = 680.15K
Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)
ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4
7.831 = Ea(2.417 × 10^-5)
Ea = 3.24 × 10^5 J/mol
The calculation for the amount of water present in the given amount of hydrate is shown below,
amount water = (100 g hydrate) x (0.347 g H2O / 0.946 g hydrate)
= 36.68 g
Thus, the amount of water present in the hydrate is approximately 36.68 g.
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
![q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5B%5Ctext%20%7Bglucose%206-phosphate%7D%5D%5BADP%5D%7D%7B%5BGlucose%5D%5BATP%5D%7D)
so,
⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
