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2 years ago

The element thallium is 70% thallium-205 and 30% thallium-203. Calculate its relative atomic mass to 1 decimal place.

Chemistry

1 answer:

xxMikexx [17]2 years ago

6 0

Answer:

answer-

The relative atomic mass = 204.4

explanation:

Thallium -203 = 30%

Thallium -205 = 70%

Therefore ,

relative mass of thallium = (30×203 + 70×205)/100

relative mass of thallium = (20440)/100

relative mass of thallium = 204.40 amu

Thus,

relative atomic mass of thalium =204.4 ( to 1 decimal place)

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How many O2 molecules are present in 0.470 g of oxygen

adoni [48]

The number of molecules in 16 grams of oxygen gas is 3.01*10^23, equal to half of Avogadro's number. ... How many molecules of oxygen (o2) are present in 16.0 g of o2…

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2 years ago

A 250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of h

Ksenya-84 [330]

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution : 0.0328 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0328=\frac{\text{Number of moles}}{0.250 L}$

Moles of HCl in 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution : 0.0245 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0245=\frac{\text{Number of moles}}{0.100 L}$

Moles of NaOH in 0.100 L solution = 0.00245 moles

3) Concentration of hydrochloric acid in the resulting solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.

Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L

Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}$

Molarity of HCl left un-neutralized : $\frac{0.00575 moles}{0.350L}=0.0164 M$

0.0164 molar concentration of hydrochloric acid in the resulting solution.

3 0

2 years ago

What is the ph of a solution made by combining 157 ml of 0.35 m nac2h3o2 with 139 ml of 0.46 m hc2h3o2? the ka of acetic acid is

ExtremeBDS [4]

The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar

Then, we calculate the pH as follows:
pKa of acetic acid = -log(1.75 × 10^-5) = 4.7569
pH = pKa + log ([salt] / [acid])
= 4.7569 + log(0.1856 / 0.216)
= 4.691

6 0

2 years ago

The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T

kifflom [539]

Answer : The equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

First we have to calculate the standard free energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]$

where,

$\Delta G^o$ = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]$

$\Delta G^o=-90.72kJ/mol$

Now we have to calculate the value of $K_p$

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G_^o$ = standard Gibbs free energy = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p$

$K_p=7.98\times 10^{15}$

Now we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium $SO_2$ pressure be, x

Pressure of $SO_2$ = Pressure of $H_2S$ = x

Now put all the given values in this expression, we get

$7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}$

$x=3.93\times 10^{-5}torr$

Thus, the equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

4 0

2 years ago

Obtain a box of breakfast cereal and read the list of ingredients. What are four chemicals from the list

FromTheMoon [43]

Options:

monoglycerides

cocamide DEA

folic acid

iron chromium ion

peroxide

lauryl glucoside

disodium phosphate

Answer and Explanation:

The added chemicals are:

monoglycerides
folic acid
iron
disodium phophates

Monoglycerides are fats added for flavour. Folic cid and iron are vitamins added for nutritional value. disodium phosphate is a food additive for enhancing flavour.

The remaining ingredients are organic based.

3 0

2 years ago