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vichka [17]
2 years ago
12

What are two possible ways to show the structure of ch4?

Chemistry
2 answers:
ikadub [295]2 years ago
8 0

Answer: Two possible ways to show the structure of CH4 are its electron dot diagram or structural formula. CH4 or methane's molecular formula is given as CH4. The structural formula is a graphical representation of a chemical compound.

Alenkasestr [34]2 years ago
7 0

Answer:

using electron dot diagrams or structional formula

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Rock samples, taken from the Moon, contain many of the same minerals as those found on Earth. Why might this be the case?
Feliz [49]
The moon used to be connected with Earth. A meteor hit several billion years ago and split the Earth and Moon apart. This explains why they share the Sam properties.
8 0
3 years ago
A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen
andrew-mc [135]
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O 
the limiting reactant in this reaction is C₄H₁₀  This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
7 0
4 years ago
If you start with 13 mole H2O and 50 grams of O2 what is the percent yield of H202 if 100 grams of H202 is actually made?
dmitriy555 [2]

Answer:

94.1 %

Explanation:

We firstly determine the equation:

2H₂O + O₂ →  2H₂O₂

2 moles of water react to 1 mol of oxygen in order to produce 2 moles of oxygen peroxide.

We convert the mass of oxygen to moles:50 g . 1mol /32g = 1.56 mol

Certainly oxygen is the limiting reactant.

2 moles of water react to 1 mol of oxygen.

13 moles of water may react to 13/2 = 6.5 moles. (And we only have 1.56)

As we determine the limiting reactant we continue to the products:

1 mol of O₂ can produce 2 moles of H₂O₂

Then 1.56 moles of O₂ will produce (1.56 . 2) = 3.125 moles

We convert the moles to mass: 3.125 mol . 34 g/mol= 106.25 g

That's the 100% yield or it can be called theoretical yield.

Percent yield = (Yield produced / Theoretical yield) . 100

(100g / 106.25 g) . 100 = 94.1 %

3 0
3 years ago
Which of the following are strong electrolytes in water?
m_a_m_a [10]

Answer:

Ionic compounds that ionize completely in water

8 0
3 years ago
Read 2 more answers
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