steel is formed when a little bit of carbon is added to which element (the carbon strengthens and hardens the steel)

An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is

Juli2301 [7.4K]

Answer:

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

Explanation:

The gas ideal law is

PV= nRT (equation 1)

Where:

P = pressure

R = gas constant

T = temperature

n= moles of substance

V = volume

Working with equation 1 we can get

$n =\frac{PV}{RT}$

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

$\frac{m}{mw} =\frac{PV}{RT}$ or

$m =\frac{P*V*mw}{R*T}$ (equation 2)

The cylindrical container has a constant pressure p

The volume is the volume of a cylinder this is

$V =(pi)*r^{2}*h$

Where:

r = radius

h = height

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is $V =(pi)*r^{2}*h$

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:

$T =\frac{T_{1} + T_{O}}{2}$

Replacing these values in the equation 2 we get:

$m =\frac{P*V*mw}{R*T}$ (equation 2)

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

8 0

3 years ago

If a substance melts at 20oC, what is its freezing point?

shutvik [7]

Answer:

It is difficult, if not impossible, to heat a solid above its melting point because the heat that ... in a solid are packed in a regular structure that is characteristic of that particular substance.

<h3>#carryONlearning </h3>

4 0

2 years ago

1. Which of the following nuclear equations is correctly balanced? (choices attached)

lawyer [7]

1. Balancing nuclear equations

Answer:

$_{18}^{37}\text{Ar} + _{-1}^{0}\text{e} \rightarrow _{17}^{37}\text{Cl}$

Explanation:

The main point to remember in balancing nuclear equations is that the sums of the superscripts (the mass numbers) and the subscripts (the nuclear charges) must balance.

Mass numbers: 37 + 0 = 37; balanced.

Charges: 18 + 1 = 17; balanced

B is wrong. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.

C is wrong. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).

D is wrong. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.

===============

2. Amount remaining

Answer:

D. 5.25 g

Explanation:

The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:

No. of Fraction Amount

half-lives t/(da) remaining remaining/g

1 24 ½ 21.0

2 48 ¼ 10.5

3 72 ⅛ 5.25

4 96 ¹/₁₆ 2.62

We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.

===============

3. Calculating the half-life

Answer:

a. 2.6 min

Explanation:

The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.

We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.

2 half-lives = 5.2 min Divide both sides by 2

1 half-life = 5.2 min/2 = 2.6 min

7 0

2 years ago

What is the mass ratio of aluminum to oxygen in aluminum oxide, Al2O3?

liubo4ka [24]

Answer:

9 : 8

Explanation:

Aluminum oxide has the following formula Al₂O₃.

Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:

Mass of Al in Al₂O₃ = 2 × 27 = 54 g

Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g

Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:

Mass of Al = 54 g

Mass of O₂ = 48 g

Mass of Al : Mass of O₂ = 54 : 48

Mass of Al : Mass of O₂ = 54 / 48

Mass of Al : Mass of O₂ = 9 / 8

Mass of Al : Mass of O₂ = 9 : 8

Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8

5 0

2 years ago

The cell potential of the following electrochemical cell depends on the pH of the solution in the anode half-cell:Pt(s)|H2(g, 1a

meriva

Answer:

0.51

Explanation:

Given the Nernst equation;

E= E° - 0.0592/n logQ

E= 355 mV or 0.355 V

E° = 0.34 - 0= 0.34 V

n= 2(two electrons were transferred in the process)

Equation of the reaction;

H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)

Substituting values;

0.355 = 0.34 - 0.0592/2 log([H^+]/1)

0.355 - 0.34 = - 0.0296 log [H^+]

0.015/-0.0296 = log [H^+]

Antilog (-0.5068) = [H^+]

[H^+] = 0.311 M

pH = -log[H^+]

pH= - log(0.311 M)

pH = 0.51

6 0

3 years ago