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3 years ago

Consider the chemical equation CH4 + 2 O2 → CO2 + 2 H2O. In this equation, CH4 is a

Physics

2 answers:

Reika [66]3 years ago

6 0

Answer:

$\huge \boxed{\mathrm{B. \ reactant}}$

$\rule[225]{225}{2}$

Explanation:

$\sf CH_4+ O_2 \Rightarrow CO_2 + H_2O$

Balancing the Hydrogen atoms on the right side,

$\sf CH_4+ O_2 \Rightarrow CO_2 +2 H_2O$

Balancing the Oxygen atoms on the left side,

$\sf CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O$

Reactants are on the left side:

$\sf \boxed{\sf CH_4 + O_2} \Rightarrow CO_2 + H_2O$

$\rule[225]{225}{2}$

Dennis_Churaev [7]3 years ago

3 0

$CH4+2O2=\ \textgreater \ CO2+2H2O$
Since terms which are placed left to the arrow mark are reactants and those placed right are products you can easily solve the task represented above. According to the information I've provided CH4 is reactant and it is Fule. So, the correct answer is B.

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How far from a converging lens with a focal length of 16 cm should an object be placed to produce a real image which is the same

Feliz [49]

Answer:

32 cm

Explanation:

f = focal length of the converging lens = 16 cm

Since the lens produce the image with same size as object, magnification is given as

m = magnification = - 1

p = distance of the object from the lens

q = distance of the image from the lens

magnification is given as

m = - q/p

- 1 = - q/p

q = p eq-1

Using the lens equation, we get

1/p + 1/q = 1/f

using eq-1

1/p + 1/p = 1/16

p = 32 cm

4 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

How do quantum numbers relate to electrons? please explain?

alukav5142 [94]

Quantum numbers allow us to both simplify and dig deeper into electron configurations. Electron configurations allow us to identify energy level, subshell, and the number of electrons in those locations. If you choose to go a bit further, you can also add in x,y, or z subscripts to describe the exact orbital of those subshells (for example 2px). Simply put, electron configurations are more focused on location of electrons then anything else.

Quantum numbers allow us to dig deeper into the electron configurations by allowing us to focus on electrons' quantum nature. This includes such properties as principle energy (size) (n), magnitude of angular momentum (shape) (l), orientation in space (m), and the spinning nature of the electron. In terms of connecting quantum numbers back to electron configurations, n is related to the energy level, l is related to the subshell, m is related to the orbital, and s is due to Pauli Exclusion Principle.

7 0

3 years ago

a hammer drops from a height of 8 meters. calculate the speed with which it hits the ground. show work

ioda

Answer:

12.5 m/s

Explanation:

The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$