How many atoms of each element are in the following compound: C3H5(NO3)3

How do you find the mass of original KClO3 sample

emmasim [6.3K]

You cannot find it if you do not know the information. Was there a molar mass given or do you know if it is asking for the molar mass?

5 0

3 years ago

The following information is given for ether, C2H5OC2H5, at 1atm: boiling point = 34.6 °C Hvap(34.6 °C) = 26.5 kJ/mol specific h

Likurg_2 [28]

Answer:

The heat is $H= -8.044KJ$

Explanation:

From the question we are told that

The pressure is $P = 1 \ atm$

The boiling point is $B_P = 34^oC$

The heat of vaporization at 34°C is = $26.5 kJ/mol$

The specific heat of the liquid is $c_p = 2.32 J/g^oC$

The mass is $m = 22.5g$

The no of moles of the sample of $C_2 H_5OC_2H_5$ is given as

$No \ mole (n) = \frac{Mass \ of \ sample }{Molar \ mass }$

The Molar mass for $C_2 H_5OC_2H_5$ is a value = $= 74.12 g/mol$

Substituting the value into the above equation

$n = \frac{22.5}{74.12}$

$= 0.30356 \ mol$

The heat H is mathematically as

$H =- nH_{vap}$

The negative sign show that the heat is for condensing

$H = 0.30356 * 26.5 *10^{3} J/mol$

$H= -8.044KJ$

6 0

4 years ago

The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance b

SSSSS [86.1K]

Answer:

the stronger light 5.5 m apart from the total illumination

Explanation:

From the problem's statement , the following equation can be deducted:

I= k/r²

where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality

denoting 1 as the stronger light and 2 as the weaker light

I₁= k/r₁²

I₂= k/r₂²

dividing both equations

I₂/I₁ = r₁²/r₂²=(r₁/r₂)²

solving for r₁

r₁ = r₂ * √(I₂/I₁)

since we are on the line between the two light sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus

r₂ = r₁ + d

then

r₁ = (r₁ + d)* √(I₂/I₁)

r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)

r₁*(1-√(I₂/I₁)) = d*√(I₂/I₁)

r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁)) =

r₁ = d/[√(I₁/I₂)-1)]

since the stronger light is 9 times more intense than the weaker

I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3

then since d=11 m

r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m

r₁ = 5.5 m

therefore the stronger light 5.5 m apart from the total illumination

5 0

3 years ago

Why is U-235 the most common nuclear fuel instead of the more abundant U-238? State a clear topic sentence and support it with e

GREYUIT [131]

<h3>Answer and explanation;</h3>

The isotope U-235 is an important common nuclear fuel because under certain conditions it can readily be split, yielding a lot of energy. It is therefore said to be 'fissile' and use the expression 'nuclear fission'.
Uranium 238 on the other hand is not fissionable by thermal neutrons, but it can undergo fission from fast or high energy neutrons. Hence it is not fissile, but it is fissionable.
In a nuclear power station fissioning of uranium atoms replaces the burning of coal or gas. Heat created by splitting the U-235 atoms is then used to make steam which spins a turbine to drive a generator, producing electricity.

3 0

3 years ago

What do they all have the same of ?

AlekseyPX

Answer:

Same # of protons.

Explanation:

It's right cuz I took the test like last year and I still remember the answer

4 0

3 years ago