Find oxidation numbers for the following: Cl: Be: Al: Ca: O:

What are the number of protons, electrons, and neutrons in one atom of titanium?.

Olin [163]

Answer:

Protons: 22

Neutrons: 26
Electrons: 22

Explanation:

Titanium's atomic number is 22, meaning that the number of protons it has in its nucleus is also 22. # of electrons = # of protons, so the number of electrons is also 22. Titanium has an atomic mass of 47.867, but we can round up to 48. To find the atomic mass, we add up number of protons and neutrons. We know the number of protons as 22, so we subtract 22 from 48 and get a number of 26. Thus, the number of neutrons is 26.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

3 0

3 years ago

What is the total number of moles of aluminum oxide that can be formed when 54 grams of aluminum reacts completely with oxygen?

umka2103 [35]

4Al + 3O₂ → 2Al₂O₃

m(Al)=54 g
M(Al)=27 g/mol

n(Al₂O₃)=n(Al)/2

n(Al)=m(Al)/M(Al)

n(Al₂O₃)=m(Al)/{2M(Al)}

n(Al₂O₃)=54/{2*27}= 1 mol

3 0

3 years ago

Read 2 more answers

riddle time We are two separate things but very much alike. We can be life long or the lesson of your life. We grant you happine

Eva8 [605]

Choices/ Decisions. Hope this helps.

8 0

3 years ago

Describe the motion of particles in the three common states of matter

stellarik [79]

Fast, slow, Non moving?

8 0

3 years ago

Read 2 more answers

Grey Goose ® vodka has an alcohol content of 40.0 % (v/v). Assuming that vodka is composed of only ethanol and water answer the

Vitek1552 [10]

Explanation:

Grey Goose vodka has an alcohol content of 40.0 % (v/v).

Volume of vodka = V = 100 mL

This means that 40.0 mL of alcohol is present 100 mL of vodka.

Volume of ethanol=V' = 40.0 mL

Mass of ethanol = m

Density of the ethanol = d = 0.789 g/mL

$m=d\times V' = 0.789 g/ml\times 40.0 mL=31.56 g$

Volume of water = V''= 100 ml - 40.0 mL = 60.0 mL

Mass of water = m'

Density of the water = d' = 1.00 g/mL

$m'=d'\times V'' = 1.00 g/ml\times 60.0 mL=60.0 g$

a.)

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

Volume of vodka = V = 100 mL = 0.100 L ( 1mL=0.001 L)

Molarity of the ethanol:

$=\frac{0.6861 mol}{0.100 L}=6.861 M$

6.861 M the molarity of ethanol in this vodka.

b) Mass of ethanol = 31.56 g

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

Volume of vodka = V = 100 mL

Mass of vodka = m

Density of the water = D = 0.935 g/mL

$M=D\times V=0.935 g/ml\times 100 ml=93.5 g$

The percent by mass of ethanol % (m/m):

$\frac{31.56 g}{93.5 g}\times 100=33.75\%$

33.75% is the percent by mass of ethanol % (m/m) in this vodka.

c)

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

Mass of solvent that is water = 60.0 g = 0.060 kg ( 1g = 0.001 kg)

Molality of ethanol in vodka :

$m=\frac{0.6861 mol}{0.060 kg}=11.435 m$

11.435 m is the molality of ethanol in this vodka.

d)

Moles of ethanol = $n_1=\frac{31.56 g}{46g/mol}=0.6861 mol$

Moles of water = $n_2=\frac{60.0 g}{18 g/mol}=3.333 mol$

Mole fraction of ethanol = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.6861 mol}{0.6861 mol+3.333 mol}$

= 0.1707

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{3.3333 mol}{0.6861 mol+3.333 mol}$

= 0.8290

e)

The vapor pressure of vodka = P

Mole fraction of ethanol = $\chi_1=0.1707$

Mole fraction of water = $\chi_2=0.8290$

The vapor pressures of ethanol = $p_1=45.0 Torr$

The vapor pressures of pure water = $p_2=23.8Torr$

$P=\chi_1\times p_1+\chi_2\times p_2$

$P=0.1707\times 45.0torr+0.8290\times 23.8 Torr=27.41 torr$

The vapor pressure of vodka is 27.41 Torr.

5 0

3 years ago