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Harlamova29_29 [7]
2 years ago
12

In the reaction between 15.8 g of NH3 and 22.7 g oxygen, water and nitrogen monoxide are

Chemistry
1 answer:
LenKa [72]2 years ago
4 0

Explanation:

limiting is NH3

NO:27/88g

H2O: 25/09g

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Fe,0, is an example of
Amiraneli [1.4K]

Answer:

it is the example of molecules

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2 years ago
Which allele for human blood type is recessive?
blagie [28]
O blood type is recessive hope this helps

8 0
3 years ago
Read 2 more answers
A projectile is fired with speed v0 at an angle theta from the horizontal from the horizontal as shown in the figure.
irga5000 [103]

Answer:

v₀ = √(2gH/(sin²θ)) = (sin θ)√(2gH)

v₀ = √(gR/(sin2θ))

Explanation:

An image of the artillery officer, the hill and path of motionof the projectile is attached to this solution.

Given, R, H, g and θ (theta)

Using the equations of motion, we can get the initial velocity v₀

First of, we need to resolve this motion into the vertical and horizontal axis.

The horizontal component of the initial velocity, v₀ₓ = v₀ cos θ

Vertical component of the initial velocity, v₀ᵧ = v₀ sin θ

When the projectile reaches maximum height, Velocity at max height, vₕ = 0m/s

From equations of motion,

vₕ = v₀ᵧ - gt

0 = v₀ sinθ - gt

t = v₀ sinθ/g

This is the time taken to reach maximum height. The time take to comolete the toyal flight, T = 2t = (2v₀ sinθ)/g

The maximum height to be reached, H can be calculated from the equations of motion too

H = vₕt - 0.5gt² = 0 - 0.5g((v₀ sinθ)/g)²

H = (0.5g v₀² sin²θ)/g²

H = (v₀² sin²θ)/2g

The range, or horizontal distance to be covered by the projectile, R, will be calculated using the horizontal component of the initial Velocity, v₀ₓ = v₀ cos θ, this horizontal velocity is constant all through the motion, so, no acceleration in the horizontal direction.

R = v₀ₓT =  (v₀ cos θ)((2v₀ sinθ)/g)

R = (v₀²(2cosθsinθ)/g)

2cosθsinθ = sin2θ

R = v₀²(sin2θ)/g

So, writing v₀ in terms of all the other parameters,

v₀ = √(2gH/(sin²θ)) =  (sinθ)√(2gH

v₀ = √(gR/(sin2θ))

4 0
2 years ago
A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi
attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

E^{o}  = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] =  log[ox] -  \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] -  \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 -  \frac{2x5.45x10^{-2}  }{0.0591}}\\\\

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

7 0
3 years ago
Why can't polyatomic ions ever stand alone?
Basile [38]
The prefix "poly-" means "many," and "atomic" refers to atoms, so a polyatomic ion is made of 2+ atoms covalently bonded together. These ions are charged, which is why they can never stand alone. 
7 0
2 years ago
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