Answer:
The acceleration of the car is 9,19 m/s2
Explanation:
We use the formula: F=m x a---> a=F/m
a=21,6N/ 2,35kg 1N is 1kgxm/s2
a=21,6 kg x m/s2 x 2,35 kg
a=9,191489362 m/s2
A - 1 CH4+ 1 O2 = 1 CO2+2 H2
b - 2 Al+3 Cl2 = 2 AlCl3
c - 1 CH2O+ 1 H2 = 1 CH3OH
*The ones that have the coefficient of 1 you can leave blank but if you have to put a number just put 1
Answer: It's equal to 10^(-2.3), or 0.00501 M, or 5.01 * 10^-3 moles/Liter
Explanation:
Well, pH = - log[H+]
Or, in words, pH is equal to -1 multiplied by the logarithm (base 10) of the hydrogen ion concentration.
So you have 2.3 = -log[H+]. We want to isolate the H+, so let's start simplifying the right hand side of the equation. First, we multiply both sides by -1.
-2.3=log[H+]
Now, the definition of a logarithm says that if the log (base 10) of [H+] is -2.3, then 10 raised to the -2.3 power is [H+]
So on each side of the equation, we raise 10 to the power of that side of the equation.
10^(-2.3) = 10^(log[H+])
and because 10^log cancels out...
10^(-2.3) = [H+]
Now we've solved for [H+], the hydrogen ion concentration!
Answer:
uh i think its D All of the above
Explanation:
sorry if its wrong
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
