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3 years ago

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Chemistry

1 answer:

Vanyuwa [196]3 years ago

4 0

The most appropriate answer is C !!

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Which of these compounds would most likely be found in a deposit of natural gas?

Valentin [98]

The correct answer for the given question above would be option A. The compound that would most likely be found in a deposit of natural gas is CH4 or METHANE. Methane is the main constituent of natural gas. It is<span> a colorless, odorless gas with a wide distribution in nature. Hope this answers your question.</span>

8 0

3 years ago

Read 2 more answers

Dissolve 30 g of sodium sulphate into 300 mL of water

Aneli [31]

Answer:

number of moles = 0.21120811

Explanation:

To find the number of moles, given the mass of the solute, we use the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n = number\:of\:moles\:(mol)}$

$\mathrm{m = mass\:of\:solute\:(g)}$

$\mathrm{M = molar\:mass\:of\:solute\:( \dfrac{ g }{ mol } )}$

Label the variables with the numbers in the problem:

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M =\:?\:Calculate\:the\:molar\:mass }$

The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:

Formula for finding the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

For the variables and what they mean are below for finding the molar mass of sodium sulfate:

$\mathrm{M =molar\:mass }$

$\mathrm{m =moles=2\:moles\:for\:Na\:,1\:mole\:for\:S,\:and\:4\:moles\:for\:O}$

$\mathrm{Na =sodium=22.99\:g }$

$\mathrm{S =sulfur=32.06\:g }$

$\mathrm{O =oxygen=16.00\:g }$

Plug the numbers into the formula, to find the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$M = 2 \left( 22.99 \right) +1 \left( 32.06 \right) +4 \left( 16.00 \right)$

$\mathrm{Multiply\:2\:by\:22.99\:to\:get\:45.98\:and\:1\:by\:32.06\:to\:get\:32.06}$

$\mathrm{M = 45.98+32.06+4\:(16)}$

$\mathrm{Multiply\:4\:by\:16\:to\:get\:64}$

$\mathrm{M = 45.98+32.06+64}$

$\mathrm{Add\:45.98\:and\:32.06\:to\:get\:78.04}$

$\mathrm{M = 78.04+64}$

$\mathrm{Add\:78.04\:and\:64\:to\:get\:142.04}$

$\mathrm{M = 142.04}$

Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M = 142.04\:g/mol}$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$\mathrm{n = \dfrac{ 30 }{ 142.04 }}$

$\mathrm{Divide\:142.04\:by\:30\:to\:get\:0.21120811}$

$\mathrm{n = 0.21120811}$

0.21120811 rounded gives you 0.2112

or if you did the problem without decimals

30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.

3 0

2 years ago

Convert 13.4 degrees celcius into kelvin

Alenkasestr [34]

Answer:

286.55K

Explanation:

To convert to kelvin , add 237 .15

$13.4\°C + 273.15 = 286.55K$

5 0

3 years ago

Read 2 more answers

When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.

sertanlavr [38]

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $15.8J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m$ = mass of water = 100.0 g

$\Delta T$ = change in temperature = $T_2-T_1=(32.0-23.9)=8.1^oC$

Now put all the given values in the above formula, we get:

$q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]$

$q=3513.8J=3.5138kJ$ (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole$

Now,

$\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol$

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

4 0

3 years ago

Calculate the mass in grams of each sample.<br><br> 4.68×1020 H2O2 molecules

oksian1 [2.3K]

24= y2k* - 2x H202 grams in mass

7 0

3 years ago