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Goryan [66]
3 years ago
10

How many atoms of carbon are combined to with 2.81 moles of hydrogen atoms in a sample of the compound ethane, C2H6?

Chemistry
1 answer:
puteri [66]3 years ago
5 0

Answer:

5.64×10²³ atoms C

Explanation:

Convert moles of H to moles of C:

2.81 mol H × (2 mol C / 6 mol H) = 0.937 mol C

Convert moles of C to atoms of C:

0.937 mol C × (6.02×10²³ atoms C / mol C) = 5.64×10²³ atoms C

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Bubbles are released when nitric acid is added to a potassium carbonate solution.
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6 0
2 years ago
Which of the following statements is not correct?
velikii [3]
To answer the question above, multiply the given number of moles by the molar masses.

(A)     (0.20 mole) x (32 g / 1 mole) = 6.4 grams O2
(B)     (0.75 mole) x (62 g / 1 mole) = 46.5 grams H2CO3
(C)     (3.42 moles) x (28 g / 1 mole) = 95.7 grams CO
(D)     (4.1 moles) x (29.88 g / 1 mole) = 122.508 g Li2O

The answer to the question above is letter D.
4 0
3 years ago
What is safer to drink 113 ml of HCl or 244 ml of NaOH​
skad [1K]

Answer:

bleach

Explanation:

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5 0
2 years ago
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
2 years ago
For the balanced equation shown below, how many moles of o2 will react with 0.3020 moles of co2? 2c2h5oh + 6o2 → 4co2 + 6h2o que
MariettaO [177]
The balanced reaction for combustion is as follows ;
2C₂H₅OH + 6O₂ ---> 4CO₂ + 6H₂O
the stoichiometry of C₂H₅OH to O₂ is 2:6
that means 2 mol of C₂H₅OH reacts with 6 mol of O₂.
when 1 mol of C₂H₅OH reacts with 6/2 mol of O₂,
then 0.3020 mol of C₂H₅OH reacts with - 6/2 x 0.3020
therefore number of O₂ moles reacted = 0.91 mol
6 0
3 years ago
Read 2 more answers
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