Answer:
Both plants and animals release energy from glucose in cellular respiration.
<span>By definition, summer is the portion of the year in which the hemisphere is tilted toward the sun, so that sunlight strikes the surface more directly. When it is summer in the northern hemisphere, it is winter in the southern hemisphere, and vice versa</span>
<span>C2H5
First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass.
moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles
moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles
The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule.
Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen.
moles C = 0.50899
moles H = 0.638361 * 2 = 1.276722
We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon.
total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185
7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked.
Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen.
0.50899 / 1.276722 = 0.398669
0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5.
Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is
C2H5</span>
Answer:
15.0 µm
Step-by-step explanation:
Density = mass/volume
D = m/V Multiply each side by V
DV = m Divide each side by D
V = m/D
Data:
m = 1.091 g
D = 7.28 g/cm³
l = 10.0 cm
w = 10.0 cm
Calculation:
<em>(a) Volume of foil
</em>
V = 1.091 g × (1 cm³/7.28 g)
= 0.1499 cm³
(b) <em>Thickness of foil
</em>
The foil is a rectangular solid.
V = lwh Divide each side by lw
h = V/(lw)
= 0.1499/(10 × 10)
= 1.50 × 10⁻³ cm Convert to millimetres
= 0.015 mm Convert to micrometres
= 15.0 µm
The foil is 15.0 µm thick.
Answer : (C) "Higher frequencies have larger spaces between lines".
Explanation:
In Young's experiment, the condition for constructive interference is given by :
.........(1)
n is order or number of lines observed
d is distance between slits
is the angle between the path and the line from screen to the slits.
We also know that, ![c=\nu \lambda](https://tex.z-dn.net/?f=c%3D%5Cnu%20%5Clambda)
or
![\lambda=\dfrac{c}{\nu}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bc%7D%7B%5Cnu%7D)
where,
c is the speed of light
is frequency
is wavelength
So, equation (1) turns into
![dsin\theta=n\dfrac{c}{\nu}](https://tex.z-dn.net/?f=dsin%5Ctheta%3Dn%5Cdfrac%7Bc%7D%7B%5Cnu%7D)
![\nu=\dfrac{n\ c}{d\ sin\theta}](https://tex.z-dn.net/?f=%5Cnu%3D%5Cdfrac%7Bn%5C%20c%7D%7Bd%5C%20sin%5Ctheta%7D)
So,
![\nu\propto n](https://tex.z-dn.net/?f=%5Cnu%5Cpropto%20n)
or
Higher frequencies have larger spaces between line.
So, correct option is (C).