Answer:
M=0.380 M.
Explanation:
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In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

Now, we compute the total moles of bromide:

Then, the total volume in liters:

Therefore, the concentration of total bromide is:

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Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:

The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M
Answer:
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Explanation:
It has been proven time and time again to ruin lungs, effectively shortenning ones lifespan.
The number of moles of a gas if it occupy a volume of 4dm at 10 atm and - 200 c is 6.67 moles
calculation
by use of ideal gas equation
that is PV =nRT
where P(pressure) = 10 atm
V( volume) = 4dm^3 = 4L
n(number of moles) =?
R(gas constant) = 0.0821 L.atm/mol.K
T = -200 +273 = 73 K
make n the formula of the subject n = PV/RT
n is therefore = (10 atm x 4l)/(0.0821 L.atm/mol.K x 73 K) = 6.67 moles