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Volgvan
3 years ago
14

How many molecules are in 1.00 g of C10H14N2? 3.71 x 1021 9.76 x 1025 1.02 x 10-26 2.32 x 1022

Chemistry
1 answer:
Reika [66]3 years ago
8 0

Answer:

3.71 x 1021 (First option)

Explanation:

Let's determine the molar mass of the compound.

C₁₀H₁₄N₂ = 12. 10 + 14 .1 + 14 .2 = 162 g/mol

So 162 grams of compound are contained in 1 mol

1 g of  C₁₀H₁₄N₂ are contained in ( 1 / 162) = 0.00617 moles

Let's conver the moles into amount of molecules (mol . NA)

0.00617 . 6.02×10²³ = 3.71×10²¹ molecules

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Use the periodic table to determine how many grams of oxygen would be required to react completely with 859.0 g C2H2?
Alekssandra [29.7K]
 <span>(a) The balanced equation shows that 2 moles of water result from 5 moles of 
oxygen that reacts, so that is 0.4 times as many moles of water as O2. So if we 
have 2.50 moles of oxygen reacting, we produce 0.4 times that amount, or 
1.0 mole of water. 
(b) The molecular weight of acetylene is 26 and that of O2 is 32, so we can set 
up the proportion: 2.25 gm. / 52 = x gm. / 160 
Then x grams = (2.25)(160) / 52 = 6.923 grams of O2 are required. 
(c) The balanced equation shows that twice as many moles of CO2 are produced 
as we have acetylene reacting. If 78.0 grams of acetylene react, that is 3 moles 
so we produce twice that, or 6 moles of CO2. The molecular weight of CO2 is 44, 
so we have 44 times 6, or 264 grams of carbon dioxide produced. 
(d) If we collect 186 grams of CO2, the percentage yield is: 
186/264 = 0.7045, or 70.45 percent yield. 
Whew! Hope this answers all parts of your question!</span><span>
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6 0
3 years ago
Read 2 more answers
A houseplant is placed on a window sill that gets direct, bright sunlight
Molodets [167]

Answer:

D. Phototropism

Explanation:

Plants, like every other living organism, responds to external stimulus. In this question, the stimulus is LIGHT from the sun entering through the window sill. The plant is responding positively to the light stimuli by growing towards the direction where the light is coming from. This exhibit is called PHOTOTROPISM.

Phototropism, comes from two words viz photo- meaning light and tropism- meaning movement. Hence, phototropism is the growth of an organism (plant in this case) in response to a light stimulus. The case described in this question is POSITIVE PHOTOTROPISM because the growth occurs towards the light source.

7 0
2 years ago
Yo yuh guys know this
erastovalidia [21]

Answer:80KM is distance. 30KM north is displacement.

Explanation:

8 0
2 years ago
Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml.
OleMash [197]

Answer: -

15.55 M

35.325 molal

Explanation: -

Let the volume of the solution be 1000 mL.

Density of nitric acid = 1.42 g/ mL

Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid

= 1000 mL x 1.42 g/ mL

= 1420 g.

Percentage of HNO₃ = 69%

Amount of HNO₃ = \frac{69} {100} x 1420 g

= 979.8 g

Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol

Number of moles of HNO₃ = \frac{979.8 g}{63 g/ mol}

= 15.55 mol

Molarity is defined as number of moles per 1000 mL

We had taken 1000 mL as volume and found it to contain 15.55 moles.

Molarity of HNO₃ = 15.55 M

Mass of water = Total mass of nitric acid solution - mass of nitric acid

= 1420 - 979.8

= 440.2 g

So we see that 440.2 g of water contains 15.55 moles of HNO₃

Molality is defined as number of moles of HNO₃ present per 1000 g of water.

Molality of HNO₃ = \frac{15.55 x 1000}{440.2}

= 35.325 molal

3 0
3 years ago
Use standard enthalpies of formation to calculate ΔH∘rxn for the following reaction______.2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) ΔH∘rxn
Ahat [919]

Answer:

-1,103.39KJ/mol

Explanation:

We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.

In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.

The standard enthalpies of the molecules above are as follows:

H2S = -20.63KJ/mol

H2O = -285.8KJ/mol

SO2 = -296.84KJ/mol

O2 = 0KJ/mol

ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3

 ΔfH⦵(O2)]

ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]

-[ 3 × -20.63)]

= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol

6 0
3 years ago
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