a) in this we are diluting a stock solution, so we can use the dilution formula
c1v1 = c2v2
where c1 is concentration and v1 is volume of the stock solution
c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting the values
6.0 M x V = 0.500 M x 110 mL
V = 9.17 mL
9.17 mL of the stock solution should be taken and diluted upto 110 mL to prepare the 0.500 M solution
b)
In this question we are given the volume taken from the stock solution , we have to find the concentration of the diluted solution
again we use the dilution formula, c1v1 = c2v2
substituting the values
6.0 M x 10.0 mL = C x 250 mL
C = 0.24 M
the concentration of diluted solution is 0.24 M
To prepare 110 ml of 0.500 M HNO3 from a 6.0 M HNO3 solution, 9.17 ml of the stock solution would have to be used. If 10.0 ml of the stock solution is diluted to a final volume of 0.250 L, the concentration of the diluted solution will be 0.24 M.
(a) In order to prepare 110 ml of 0.500 M HNO3 from a 6.0 M HNO3 solution, we have to use the formula M1V1 = M2V2 where M and V are the molarity and volume respectively. Here, the M1 and V1 are the molarity and volume of the stock solution and M2 and V2 are the molarity and volume of the diluted solution. Filling in known values, 6.0M * V1 = 0.500M * 110ml. Solving for V1, we get V1 = (0.500 M * 110 ml) / 6.0 M = 9.17 ml. So, you would have to use 9.17 ml of the stock solution.
(b) The diluted solution's molarity is calculated using the same formula as before. Substituting the known values 6.0M * 10.0 ml = M2 * 0.250 L, rearrange the formula to get M2= (6.0M * 10.0 ml) / 0.250 L = 0.24 M or 240 mM. Therefore, the concentration of the diluted solution is 0.24 M.
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Answer: The resulting mass would be 35 grams.
Explanation: Mass cannot be destroyed, there would be no need to re-measure the mass. Therefore, if we add the masses we get 35.
Answer:
8.46 liters of 0.123 NaOH contains 25.0 grams of NaOH.
Explanation:
Molar is defined as Moles/Liter. 0.123 M means there is 0.123 moles of NaOH per 1 liter. We are asked to determine how many milliliters of this concentration NaOH it would take to deliver 25.0 grams of NaOH.
Use the units to guide the calculation.
Remember that: Concentration x Volume = Moles
(0.123 moles/liter)*(V) = moles
We want 25.0 grams of NaOH. Let's convert that to moles using the molar mass of NaOH:
(25.0 grams NaOH)/(24.0 g/mole) = 1.04 moles NaOH.
Now we can write:
(0.123 moles/liter)*(V) = 1.04 moles NaOH
V = (1.04/0.123) Liters
V = 8.46 liters
Answer:
See explanation
Explanation:
The question is incomplete. However, base on the given information, one likely question could be that, we calculate the the volume of the nitrogen gas after heating it.
So, we have that:
Initial Volume, V1 = 200mL
Initial Temperature, T1 = 24k
Final Temperature, T2 = 47k
To solve the final volume of the gas, we apply the following ideal gas law equation
V1 * T1 = V2 * T2
Substitute values for V1, T1 and T2
200mL * 24K = V2 * 47K
200mL * 24 = V2 * 47
4800mL = 47V2
Solve for V2
V2 = 4800/47
V2 = 102.13 mL ------- Approximated
Hence, the volume of nitrogen gas after heating is 102.13mL
B. nuclear separation.
C. nuclear fission.
D. nuclear fusion.
Answer is: C. nuclear fission.
Nuclear fission is a nuclear reaction or a radioactive decay where nucleus of atom split into smaller ligher nuclei.
Nuclear fission is exothermic reaction which release large amounts of energy (electromagnetic radiation or as kinetic energy, which heat reactors where fission reaction take place).
Answer: The process of splitting of an atom into two lighter atoms is called is called nuclear fission.
Explanation:
There are two types of nuclear decay reactions:
Hence, the process of splitting of an atom into two lighter atoms is called is called nuclear fission.
B. the amount of matter in a substance
C. twice the amount of mass of a substance
D. the amount of space a substance takes up
The answer is D.
Hope this helps.
b. will be easily separated.
c. lose their own properties.