A chemist prepares a solution of magnesium fluoride MgF2 by measuring out 0.00598μmol of magnesium fluoride into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's magnesium fluoride solution. Round your answer to 2 significant digits.
Answers
Answer:
0,12 μmol/L of MgF₂
Explanation:
Preparation of solutions is a common work in chemist's life.
In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in μmol/L
You have 0,00598 μmol but not Liters.
To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:
50,0 mL (1L/1000mL) = 0,05 L of water.
Thus, concentration in μmol/L is:
0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-
I hope it helps!
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Ο Ο Ο Ο
O get paper towels
O change lab stations
turn off the burner
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B. Rh
C. Hf
D. Co
Answers
Answer: C
Explanation:
Answers
Answer:
54%
Explanation:
The balanced equation is:
The first step is to determine the limiting reactant. For this, we calculate the moles of each given component and divide the result for the stoichiometric coefficient.
3.4 g octane / 114.23 g/mol = 0.030 mol octane
0.030 mol octane/2=0.015
5.9 g O2 / 32 g/mol = 0.18 mol O2
0.18 mol O2/25= 0.0074 mol
The lower number, in this case oxygen, is the limiting reactant. The value corresponds to the theoretical yield of the reaction.
Similarly, the real yield is calculated from the product.
2.80 g CO2/ 44.01 g/mol = 0.0636 mol CO2
0.0636 mol CO2/16 = 0.00398 mol
The percent yield is the ratio of the 2 multiplied by a hundred, then
Percent yield= 0.0398/0.0074 *100 = 54%
Final answer:
The percent yield of carbon dioxide from the combustion reaction of octane and oxygen, given the provided masses of the reactants and the yield of CO2, is calculated to be 26.7%
Explanation:
The combustion of octane in oxygen yields carbon dioxide and water in a 1:1 ratio, as the balanced chemical equation for the reaction is 2C8H18 + 25O2 -> 16CO2 + 18H2O. To calculate the percent yield of CO2, we first need to determine the theoretical yield of CO2. We can use the provided masses of octane and O2 and their respective molar masses to calculate the number of moles of each reactant:
- Moles of octane = 3.4 g / 114.22 g/mol = 0.0298 mol
- Moles of O2 = 5.9 g / 32.00 g/mol = 0.184 mol
Now, using the stoichiometric relationship from the balanced chemical equation, we can calculate the theoretical yield of CO2:
- Theoretical yield = 0.0298 mol octane x 16 mol CO2/2 mol octane = 0.238 mol CO2
Next, we convert this to grams using the molar mass of carbon dioxide:
- Theoretical yield = 0.238 mol CO2 x 44.01 g/mol = 10.5g CO2
Now that we have both the actual yield (2.80 g) and the theoretical yield (10.5 g), we can calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) x 100% = (2.80 g / 10.5 g) x 100% = 26.7%
Learn more about Percent Yield here:
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of this experiment?
Xe(g) + 2 F2 (g) - XeF. (g)
Answers
The percentage yield of XeF from the concentration of the given reactants will be 25%.
The reaction states that 1 mole of Xe will give 1 mole of XeF, and 2 moles of fluorine, will gives 1 mole of XeF.
The limiting reactant can be calculated:
1 mole Xe = 2 moles of
2.2 moles of Xe = 2.2 2 moles of
2.2 moles of Xe = 4.4 moles of
Since the amount of available has been in the limiting, thus
has been the limiting reactant.
So, the yield of XeF in terms of will be:
2 moles = 1 mole XeF
Thus the theoretical yield of XeF is 1 mole.
The yield of XeF we get = 0.25 moles.
Thus the percentage yield =
Percentage yield =
Percentage yield = 25%
Thus the percentage yield of XeF from the concentration of the given reactants will be 25%.
For more information about the percent yield, refer to the link:
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4
Answers
Answer:
B. NH3 < PH3< CH4
Explanation:
Hello,
In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C, -161.5 °C and -87.7 °C, clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:
B. NH3 < PH3< CH4
Best regards.
the length of ribbon needed to tie around a vase
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the size of a student's waist
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D.
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Answers
Answer:
C
the distance from the ground to the top of a ramp