Calculate the volume in milliliters of a iron(II) bromide solution that contains of iron(II) bromide . Round your answer to sign
1 answer:
This is an incomplete question, here is a complete question.
Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.
Answer : The volume of iron(II) bromide solution is,
Explanation : Given,
Concentration of iron(II) bromide = 1.29 mo/L
Moles of iron(II) bromide = 275 mmol = 0.275 mol
conversion used : 1 mmol = 0.001 mol
Now we have to calculate the volume of iron(II) bromide.
Now put all the given values in this formula, we get:
Thus, the volume of iron(II) bromide solution is,
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Charles law gives the relationship between temperature and volume of gases. It states that the volume of gas is directly proportional to temperature at constant pressure.
V / T = k
where V - volume and T - temperature in Kelvin and k - constant
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - 20 °C + 273 = 293 K
substituting these values in the equation
T = 586 K
temperature in celsius = 586 K - 273 = 313 °C
new temperature is 313 °C
V / T = k
where V - volume and T - temperature in Kelvin and k - constant
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - 20 °C + 273 = 293 K
substituting these values in the equation
T = 586 K
temperature in celsius = 586 K - 273 = 313 °C
new temperature is 313 °C
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose =
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution =
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution =
Volume of the solution taken =
Molarity of the solution after dilution =
Volume of the solution after dilution=
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose =
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Answer:
The drawing of the structure is found in diagram 1 of the attached figure.
Explanation:
Diagram 1 shows that three different types of protons are found in the structure. The nine hydrogen atoms have a similar behavior, the six hydrogen atoms also have a similar behavior and finally, the three hydrogen atoms adjacent to oxygen have a similar behavior. The number of peaks are as follows:
9H = singlet peak = between 3 and 4 ppm
6H = singlet peak = 4 ppm
3H = singlet peak = 3 ppm.
The 9 protons are around 3.5 ppm and the 6 hydrogen atoms show a peak at 4 ppm, and finally, the 3 protons have a peak around 3 ppm. Therefore, the corresponding drawing can be seen in diagram 2.
