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3 years ago

This term ____ Location refers to using another location as a reference point, rather than using latitude and longitude.

1 answer:

4 0

This term absolute location refers to using another location as a reference point, rather than using latitude and longitude.

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Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressur

Natalka [10]

Answer: a. 17.7 KJ/Mol

b. T=210K

Explanation:

Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressure is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate.

the question isnt completely originally, but we could look at the likely derivation from the questions

(a) the standard enthalpy of vaporization

using the clausius clapeyron equation

In (PT1vap / PT2vap) = delta H (vap) / R ( (1/T1) - (1/T2) )

In (35Torr/253Torr) = delta H (vap) / 8.3145 ( (1/189.55) - (1/161.2) )

Therefore, Delta H (vap) = 17.7 KJ/Mol

b. Also the boiling point

What is the normal boiling point of arsine?

At the boiling point Pvap = atmospheric pressure = 1 atm=760 torr

substitution into the equation as stated in question 1

ln(760/253)=17700/8.314(1/189.55-1/T)

T=210K

5 0

3 years ago

Which of the following is an example of heat being transferred by radiation? Choose all that apply, Fire, Sun, Candle Flame, Sto

OlgaM077 [116]

Answer:

Fire, Sun, Candle Flame, Stove

4 0

2 years ago

Read 2 more answers

3.Three resistors of 25.0Ω, 30.0Ω, and 40.0Ω are in a series circuit with a 6.0-volt battery. What is the current in the circuit

AURORKA [14]

Answer:

Current = 0.063 Amperes

Explanation:

Let the three resistors be R1, R2, and R3 respectively.

Given the following data;

R1 = 25.0Ω,

R2 = 30.0Ω

R3 = 40.0Ω

Voltage = 6 Volts

First of all, we would determine the equivalent or total resistance;

Total resistance (in series) = R1 + R2 + R3

Total resistance = 25.0Ω + 30.0Ω + 40.0Ω

Total resistance = 95 Ω

Next, we find the current flowing through the circuit;

Voltage = current * resistance

Substituting into the formula, we have;

6 = current * 95

Current = 6/95

Current = 0.063 Amperes

5 0

3 years ago

What is the equation for finding the acceleration of an object moving in a straight line?

lapo4ka [179]

The equations to find the acceleration are the suvat equations:

$v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as\\s=(\frac{u+v}{2})t$

Explanation:

The acceleration of an object is the rate if change in velocity of the object. It is calculated as

$a=\frac{v-u}{t}$

where

v is the final velocity of the object

u is the initial velocity

t is the time elapsed

For an object moving in a straight line at constant acceleration, there are several equations that can be used to find the acceleration: they are called suvat equations. They are the following:

$v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as\\s=(\frac{u+v}{2})t$

where

u is the initial velocity

v is the final velocity

t is the time

s is the distance covered

a is the acceleration

Therefore, any of the above equations can be used to calculate the acceleration.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

5 0

2 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

11 months ago