Answer:
375 grams of 4% NaOH solution contains 15 grams NaOH
Explanation:
Given 4% solution of NaOH = 4g NaOH/100 g Solution
=> 4g NaOH/100g Soln = 15 g NaOH / X
=> X = 15 g NaOH (100 g Soln) / 4 g NaOH = 375 grams Soln contains 15 g NaOH.
These questions are all about indirect and direct variation with PV=nRT, the ideal gas equation
Q3.
false, because of PV=nRT, the ideal gas equation if V goes up, P has to go down to equal nRT
Q4. false, if V remains constant, and P and T are constant as moles of gas are added, then something is wrong becse something has to change when stuff is added (V has to go down)
Q5.
PV=nRT
when T and n are constant, (R is the gas constant)
PV=k, aka V=k/P which means inversly proportional
TRUE
Q6.
ggeasy
refer to past question
PV=k
if P is doubled then V has to halve in order to equal k
1/2 times 2=1
volume is halved
Q7. use charles law
V/T=k
so
given
V=4
T= kelvins, so 299
4/299=k
so when temp goes to 22 does V go to 3.95
4/299=3.95/295?
true
because they're equal
Q8
FALSE, must be used in kelvins
T=absolute tempurature in kelvins
Q9
PV=nRT
solve for T
(PV)/(nR)=T
use final volumes and pressures
P=5atm
V=24L
n=1
R=0.082057 atm L/(mol K)
(5atm*24L)/(1mol*0.082057 atm L/mol K)=T
see, if you didn't mess up, the units cancel nicely
T=1462.4
1200 K is closest
Q10
PV/T=constant because moles are constant (supposedly)
V=4L
P=2.08atm
T=275K
so find initial to final is constant
(2.08atm*4L)/(275K)=(Pfinal*2.5L)/(323K)
solve for Pfinal
Pfinal=3.92315 atm
answer is 3.9atm
Merry Christmas
Answer:
64,433.6 Joules
Explanation:
<u>We are given</u>;
- Volume of water as 220 mL
- Initial temperature as 30°C
- Final temperature as 100°C
- Specific heat capacity of water as 4.184 J/g°C
We are required to calculate the amount of heat required to raise the temperature.
- We know that amount of heat is calculated by;
Q = mcΔT , where m is the mass, c is the specific heat, ΔT is the change in temperature.
Density of water is 1 g/mL
Thus, mass of water is 220 g
ΔT = 100°C - 30°C
= 70°C
Therefore;
Amount of heat, Q = 220g × 4.184 J/g°C × 70°C
= 64,433.6 Joules
Thus, the amount of heat required to raise the temperature of water is 64,433.6 Joules
Answer:
0.15 L
Explanation:
You need to first find the volume of the container. You can do this by dividing the mass by the density. This will give you the mass in mL.
5.00 kg = 5,000 g
(5,000 g)/(1.00 g/mL) = 5,000 mL
5,000 mL = 5 L
Now, find the volume the seawater will take up.
(5,000 g)(1.03 g/mL) = 4854.4 mL
4854.4 mL = 4.85 L
Subtract the two volumes to find the volume that left unfilled.
5 L - 4.85 L = 0.15 L
