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3 years ago

9

Which of the following is NOT a sign of a chemical reaction:

2 answers:

Sergeeva-Olga [200]3 years ago

7 0

The answer is B. Precipitation

Liono4ka [1.6K]3 years ago

3 0

I think it’s B. I apologize if I was incorrect.

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How many moles of H2SO4 are needed to prepare 200 mL of a 1.0 M solution?

jok3333 [9.3K]

The answer is 0.2moles

4 0

3 years ago

Which of the following most likely happens when the volume of a gas increases?

mafiozo [28]

the pressure of the gas increases

6 0

3 years ago

pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha

Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

* $W=1142.86Joule$

* $Q=997.7J$

* $H=2140.5J$

Explanation:

From the question we are told that:

Temperature $T=337K$

Pressure $P=(60-55)Pa*10^5$

Volume $V=(1.6-1.4)m^3*10^{-3}$

Generally the equation for gas Constant is mathematically given by

$\frac{P_2}{P_1}=\frac{V_1}{V_2}^n$

$\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n$

$n=0.65$

Therefore

Work-done

$W=\int{pdv}$

$W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}$

$W=1142.86Joule$

Generally the equation for internal energy is mathematically given by

$Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}$

$Q=997.7J$

Therefore

$H=Q+W$

$H=997.7J-11.42.9$

$H=2140.5J$

5 0

2 years ago

A runner runs for a distance of 350 yards and then runs for an additional 200 m. What is the total difference covered by the run

Rudiy27

Answer: 520.04 meters.

If you convert 350 yards to meters, you get 320.04.

320.04m + 200m = 520.04m.

Take off the .04 if necessary:)

3 0

2 years ago

he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp

wolverine [178]

1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2) $5.6\cdot 10^{13}$ electrons per person

Explanation:

1)

For a current flowing through a conductor, the drift velocity of the electrons is given by the equation:

$v_d=\frac{I}{neA}$

where

I is the current

n is the concentration of free electrons

$e=1.6\cdot 10^{-19}C$ is the electron charge

A is the cross-sectional area of the wire

The cross-sectional area can be written as

$A=\pi r^2$

where r is the radius of the wire. So the equation becomes

$v_d=\frac{I}{ne\pi r^2}$

In this problem, we have:

I = 8.0 A is the current

$8.5\cdot 10^{28} m^{-3}$ is the concentration of free electrons

d = 1.5 mm is the diameter, so the radius is

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

Therefore, the drift velocity is:

$v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s$

2)

The total length of the cord in this problem is

L = 3.00 m

While the cross-sectional area is

$A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2$

Therefore, the volume of the cord is

$V=AL$ (1)

The number of electrons per unit volume is $n$ , so the total number of electrons in this cord is

$N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}$

In total, the Earth population consists of 8 billion people, which is

$N'=8\cdot 10^9$

Therefore, the number of electrons that each person would get is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$

7 0

3 years ago