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katrin [286]
3 years ago
7

Hwlp meeeeeeeeeeeeeeeeeeee

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
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