Take the number of the exponent and move the decimal that many places. If it is positive move it left and if it is negative move it right.
For Example:
2.57x10³= 2570
Fill in any access spaces with zero.
Taking that same number to the negative 3 power
2.57x10^-3
0.00257
I put a 0. to show it is a distinctive decimal.
Answer:
Molarity = moles ÷ liters
to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)
Na = 22.989769
Br = 79.904
molar mass of NaBr = 102.893769
6.6g ÷ 102.893769 = 0.064143826 moles of NaBr
0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %
The answer is most definitely “A”
The answer to this is c i believe
Answer:
Yes, there will be liquid present and the mass is 5.19 g
Explanation:
In order to do this, we need to use the equation of an ideal gas which is:
<em>PV = nRT (1)</em>
<em>Where:</em>
<em>P: Pressure</em>
<em>V: Volume</em>
<em>n: number of moles</em>
<em>R: gas constant</em>
<em>T: Temperature</em>
we know that the pressure is 856 Torr at 300 K. So, if we want to know if there'll be any liquid present, we need to calculate the moles and mass of the CCl3F at this pressure and temperature, and then, compare it to the initial mass of 11.5 g.
From (1), solving for moles we have:
<em>n = PV/RT (2)</em>
Solving for n:
P = 856/760 = 1.13 atm
R = 0.082 L atm / mol K
n = 1.13 * 1 / 0.082 * 300
n = 0.0459 moles
Now, the mass is:
m = n * MM (3)
The molar mass of CCl3F reported is 137.37 g/mol so:
m = 0.0459 * 137.37
m = 6.31 g
Finally, this means that if we put 11.5 g of CCl3F in a container, only 6.31 g will become gaseous, so, this means it will be liquid present, and the mass is:
m = 11.5 - 6.31
m = 5.19 g