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2 years ago

The concentration of C29H60 in summer rainwater is 34 ppb. Find the molarity of this compound in nanomoles per liter (nM).

Chemistry

1 answer:

Naya [18.7K]2 years ago

5 0

Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter

Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.

The algebraic expression would be:

<em>ppb [=] micrograms of compound/liter of solution</em>

We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.

For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.

The respective procedure is in a attached file.

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What is the mass of 0.921 moles of sulfur dioxide gas (SO2)?

Serhud [2]

Answer:

mass = 58.944 g

Explanation:

Given data:

Number of moles of SO₂ = 0.921 mol

Mass of SO₂ = ?

Solution:

Formula:

Number of moles = mass/ molar mass

First of all we will calculate the molar mass.

SO₂ = 32 + 16×2 = 64 g/mol

Now we will put the values in formula.

Number of moles = mass/ molar mass

0.921 mol = mass /64 g/mol

mass = 0.921 mol × 64 g/mol

mass = 58.944 g

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2 years ago

How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K

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2 years ago

Which of the following statements best describes the Aufbau principle?

dsp73

The answer is the choice A

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2 years ago

Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l

Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]$

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol$

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0

2 years ago

the half life of the radioactive element strontium-90 is 29.1 years. If 16 grams of strontium-90 are initially present, how many

8_murik_8 [283]

Answer:

4 g after 58.2 years

0.0156 After 291 years

Explanation:

Given data:

Half-life of strontium-90 = 29.1 years

Initially present: 16g

mass present after 58.2 years =?

Mass present after 291 years =?

Solution:

Formula:

how much mass remains =1/ 2n (original mass) ……… (1)

Where “n” is the number of half lives

to find n

For 58.2 years

n = 58.2 years /29.1 years

n= 2

or 291 years

n = 291 years /29.1 years

n= 10

Put values in equation (1)

Mass after 58.2 years

mass remains =1/ 22 (16g)

mass remains =1/ 4 (16g)

mass remains = 4g

Mass after 58.2 years

mass remains =1/ 210 (16g)

mass remains =1/ 1024 (16g)

mass remains = 0.0156g

8 0

2 years ago

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