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3 years ago

15

Is jewelry gold a compound or mixture?

2 answers:

USPshnik [31]3 years ago

7 0

I think it’s a mixture.

loris [4]3 years ago

4 0

Answer:

jewelry gold is a mixture.

Explanation:

it contains more than one element and its composition varies.

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Which part of the electromagnetic spectrum have the highest level of photon energy

Pani-rosa [81]

The energy carried by one photon is directly proportional to its
frequency. So the photon energy is greatest for the electromagnetic
waves with the highest frequency / shortest wavelengths.

That's why when you get past visible light and on up through ultraviolet,
X-rays, and gamma rays, the radiation becomes dangerous ==> each
photon carries enough energy to tear electrons away from their atoms,
ripping molecules apart and damaging cells.

The photon with the highest energy is a gamma-ray photon.

4 0

3 years ago

The potential difference in a simple circuit is 2 v and the resistance is 2 ω . what current flows in the circuit? answer in uni

balandron [24]

Hey

Potential Difference given is : 2V

Resistance is : 2 ohms

By Ohm's Law, one can easily utilize the relation :

$v = ir$

Where, { v , i , r } are the potential difference, current and Resistance Respectively.

Hence,
$i = \frac{v}{r} = \frac{2}{2} = 1a$

Hence, the Current is 1 Ampere

7 0

3 years ago

A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c

tatuchka [14]

We have that the values for F north, F east, F up are

$F_N=1.09090909*10^{-5}$
$F_E=5.18181818*10^{-6}$
$F_E=2*10^{-6}$

From the Question we are told that

electric force $F_1 = 1.2 x 10^{-3} N(N)$

electric force , $F_2=5.7 x 10^{-4} N(E)$

electric force , $F_3=2.2 x 10^{-4} N (U)$

charge on this ball one $q_1= 110 nC.$

charge on this ball two $q_2= -50 nC.$

Generally the equation for the F north is mathematically given as

$F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}$

$F_N=1.09090909*10^{-5}$

For F East

$F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}$

$F_E=5.18181818*10^{-6}$

For F UP

$F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}$

$F_E=2*10^{-6}$

For more information on this visit

brainly.com/question/21811998

5 0

3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0

3 years ago

Which of the following are equivalent units?

ratelena [41]

Answer:

B is the correct answer

Explanation:

5 0

3 years ago