Ammonia , NH3, equation stocihiometric balance

: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be

Jet001 [13]

Answer: New concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.

Explanation:

The given data is as follows.

Moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 mol

Moles of $H_{2}C_{6}H_{5}O_{7}$ = 0.08 mol

Therefore, moles of $OH^{-}$ added are as follows.

Moles of $OH^{-}$ = $0.125 \times \frac{25}{1000}$

= 0.003125 mol

Now, new moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 + 0.003125

= 0.063125

Therefore, new concentration of $HC_{6}H_{5}O^{2-}_{7}$ will be calculated as follows.

Concentration = $\frac{0.063125}{0.275}$

= 0.23 M

Thus, we can conclude that new concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.

4 0

3 years ago

Question

Shalnov [3]

The Kelvin Scale is used, so the answer is temperature :)

8 0

3 years ago

Determine the mass of 1 mol of gold (Au), of aluminum chloride (AlCl3), and of glucose (C6H12O6)

sleet_krkn [62]

Answer: Mass of 1 mol of gold (Au) is 197g , of aluminum chloride $(AlCl_3)$ is 133.5 g and of glucose $(C_6H_{12}O_6)$ is 180 g

Explanation: Mass of 1 mole of a substance is called its molar mass.

a) 1 mole of $Au$ contains $6.023\times 10^{23}atoms$ and weigh 197g.

b) 1 mole of $AlCl_3$ will weigh= $1\times {\text {mass of 1 mole of Aluminium}}+3\times {\text {mass of 1 mole of chlorine atom}}$

Thus mass of 1 mole of $AlCl_3$ will be $=1moles\times {27g/mol}+3moles\times {35.5g/mol}=133.5g$

c) 1 mole of $C_6H_{12}O_6$ will weigh= $6\times {\text {mass of 1 mole of carbon}}+12\times {\text {mass of 1 mole of hydrogen atom}}+6\times {\text {mass of 1 mole of oxygen}}$

1 mole of $C_6H_{12}O_6$ will weigh= $6\times 12g/mol+12\times 1g/mol+6\times 16g/mol=180g$

5 0

4 years ago

How many milliliters of liquid does the larger graduated cylinder contain?

valina [46]

10000. millimeters. is. correct

7 0

4 years ago

Iron has many isotopes but only 4 are found in significant amounts in naturally found mixtures. The amounts by mass percent are:

AnnZ [28]

Answer:

The average mass of iron to be is 54.76 amu.

Explanation:

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

1)

Mass of Fe-54 isotope = 53.9396 amu

Percentage abundance of Fe-54 = 5.845%

Fractional abundance of Fe-54 = 0.05845

2)

Mass of Fe-56 isotope = 55.9349 amu

Percentage abundance of Fe-56 = 91.754 %

Fractional abundance of Fe-56= 0.91754

3)

Mass of Fe-57 isotope = 56.9354 amu

Percentage abundance of Fe-57 = 2.119%

Fractional abundance of Fe-57 = 0.002119

4)

Mass of Fe-58 isotope = 57.9333 amu

Percentage abundance of Fe-58 = 0.282%

Fractional abundance of Fe-58 = 0.00282

Average atomic mass of iron :

$=53.9396 amu\times 0.05845+55.9349 amu\times 0.91754 +56.9354 amu\times 0.002119 + 57.9333 amu\times 0.00282=54.76 amu$

The average mass of iron to be is 54.76 amu.

5 0

3 years ago