All categories

3 years ago

Haw does an atom change if all of it’s electrons removed

Chemistry

1 answer:

I am Lyosha [343]3 years ago

8 0

The atom becomes positively charged.
Hope this helps!

You might be interested in

9. What are the advantages of using an indicator to inform pH measurements? What are the advantages of using a pH meter?

Llana [10]

Answer:

The advantages of using an indicator to inform pH measurements:

It gives a mathematically result of the pH, in addition, it gives the precise pH of solvent, and it also gives an idea of the straight of the solution also.

Now, the advantage of using a pH meter:

It is a rapid method to characterize between acids, bases. However, this method does not show how strong acid or base actually are, plus it tends to gives a range of acidity or basicity not quite accurate as a result.

3 0

3 years ago

A 1.0 g sample of hydrogen reacts completely with 19.0 g of fluorine to form a compound of hydrogen and fluorine. a. What is the

Rashid [163]

Explanation:tr

a) Molar mass of HF = 20 g/mol

Atomic mass of hydrogen = 1 g/mol

Atomic mass of fluorine = 19 g/mol

Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{Molar mass of compound }}\times 100$

Percentage of fluorine:

$\frac{1\times 19 g/mol}{20g/mol}\times 100=95\%$

Percentage of hydrogen:

$\frac{1\times 1g/mol}{20 g/mol}\times 100=5\%$

b) Mass of hydrogen in 50 grams of HF sample.

Moles of HF = $\frac{50 g}{20 g/mol}=2.5 mol$

1 mole of HF has 1 mole of hydrogen atom.

Then 2.5 moles of HF will have:

$1\times 2.5 mol=2.5 mol$ of hydrogen atom.

Mass of 2.5 moles of hydrogen atom:

1 g/mol × 2.5 mol = 2.5 g

2.5 grams of hydrogen would be present in a 50 g sample of this compound.

c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.

Then mass of hydrogen in 50 grams of HF compound we will have :

5% of 50 grams of HF = $\frac{5}{100}\times 50 g=2.5 g$

8 0

3 years ago

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

How many valence electrons are in each atom? <br> a. potassium b. carbon c. magnesium d. oxygen

Alona [7]

A. 1 valence electron
b. 4 valence electrons
c. 2 valence electrons
d. 6 valence electrons

8 0

2 years ago

You gained two pounds while on vacation. Which of the following has changed?

Elenna [48]

I believe it is d. physical shape n particle amount (amount of particle increase if u gain weight right?), weight directly proportional to mass..

3 0

4 years ago

Read 2 more answers