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3 years ago

One of the isotopes of sulfur is represented as sulfur-35 or S-35. What does the number 35 signify

2 answers:

vladimir1956 [14]3 years ago

6 0

The number represents the total number of protons and neutrons in the nucleus also called the atomic mass.

Hope this helps! Please rate best answer if it does :)

Talja [164]3 years ago

6 0

Answer:

mass number

Explanation:

Here 35 is the mass number of sulphur. Mass number is equal to the number of protons and number of neutrons. The atomic number of S is 16. Atomic number is equal the number of protons which is equal to the number of electrons.

The electron configuration of S is 1s²2s²2p⁶3s²3p⁴

Number of neutrons in S is the difference between mass number and atomic number, which is

Number of neutrons = mass number - atomic number = 35-16 = 19

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago