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guapka [62]
3 years ago
14

For the reaction 2 H2(g) + O2(g) → 2 H2O(g), how many liters of water can be made from 5.0 L of

Chemistry
1 answer:
Lana71 [14]3 years ago
6 0

Answer: 5L O2 x 1 mol O2/ 22.4L O2 x 2 mol H20/ 1 mol O2 x 22.4L H20/ 1 mol H20 = 10 L H2O

Explanation:

For future reference though, since its at STP that means that the coefficient are in proportion. Since oxygen has a coefficient of 1 and water has a coefficient of 2 for every 1 liter of oxygen there is 2 liters of water. Hence you started with 5 liters and ended with 10

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A cell with osmolarity measuring 300 mosm/l is placed in a beaker that contains a nacl solution with osmolarity of 150mosm/l. pr
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What evidence supports a conservation law?
Murljashka [212]

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A- Energy from sunlight becomes glucose during photosynthesis

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2 years ago
Read 2 more answers
Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10
kkurt [141]

Answer:

The diameter of the hydrogen \mathbf{d =1.0605 \times 10^{-10}\ m}

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

L = \dfrac{nh}{2 \pi}

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

mvr= \dfrac{nh}{2 \pi}

v= \dfrac{nh}{2 \pi mr} ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}   ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2

ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}

r =\dfrac{n^2h^2}{4 \pi^2 mke^2}

For ground-state n = 1

h = (6.625 \times 10^{-34} \ J.s)^2

m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)

Ke = (1.6 \times 10^{-19} \ C)^2

r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}

r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}

\mathbf{r = 5.3025 \times 10^{-11} \ m}

Therefore, the diameter of hydrogen d = 2r

\mathbf{d = ( 2 \times  5.3025 \times 10^{-11} \ m})}

\mathbf{d =1.0605 \times 10^{-10}\ m}}

4 0
3 years ago
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