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Damm [24]
3 years ago
14

Which isotope is used to date ancient artifacts such as fossils?

Chemistry
1 answer:
RideAnS [48]3 years ago
6 0

Scientists use carbon-14 to date ancient fossils.

Plants and animals absorb carbon-14 produced by cosmic rays. The ratio of C-14 to C-12 is constant when they are alive.

When they die, the C-14 decays to C-12, and the ratio changes.

Carbon-14 has a half-life of 5730 years. Thus, the object loses half its C-14 every 5730 years.

Scientists can use the C-14 to C-12 ratio to date fossils up to 70 000 years old.

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The vapor pressure of water at 25.0°c is 23.8 torr. determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.
svp [43]
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958   = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass 
                                      = 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water 
                                        = 4.7 / 0.88 = 5.34 moles 
∴ moles of the solution = total moles in solu - moles of water 
                                       = 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
                                                    = 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
                                           = 4.49 g / 60 g /mol
                                           = 0.07 mol
moles of methanol = mass weight / molar mass 
                                 = 39.9  g / 32  g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg 
∴Pv(solu) = 84.55 mmHg


 
7 0
3 years ago
Determine the pH of a 2.8 ×10−4 M solution<br> of Ca(OH)2.
shepuryov [24]

Answer:

pH = 10.75

Explanation:

To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]

Using the equation:

pH = 14 - pOH

We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

pOH is

pOH = -log 5.6x10⁻⁴M

pOH = 3.25

pH = 14-pOH

<h3>pH = 10.75</h3>
3 0
3 years ago
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Klio2033 [76]

Answer:

Explanation:

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4 0
2 years ago
2.38g of black copper(iii) oxide is completely reduced to by hydrogen to give copper and water. What are the number of atoms of
OleMash [197]

Explanation:

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6 0
2 years ago
How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?
algol [13]

Answer:

1.445\times 10^{23} formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = \frac{13.0 g}{162.5 g/mol}=0.08 mol

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

3\times 0.08 mol=0.24 mol of chloride

Na^++Cl^-\rightarrow NaCl

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

\frac{1}{1}\times 0.24 mol=0.24 mol of NaCl

1 mole = N_A=6.022\times 10^{23} molecules/ atoms

Number of NaCl molecules in 0.24 moles :

=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules

1.445\times 10^{23} formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

4 0
3 years ago
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