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3 years ago

Solid iron reacts with aqueous copper(II) sulfate

Chemistry

1 answer:

Tatiana [17]3 years ago

5 0

Answer :

Fe(s) + CuSo4(aq) arrow Cu(s) +FeSo4(aq)

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Which one of these 3 ( beta, alpha, gamma) is the strongest and which is the weakest?

prisoha [69]

Answer: beta, gamma, alpha

Explanation: Beta is weakest gamma is middle alpha is strongest

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3 years ago

Ultraviolet radiation has a frequency of 6.8×1015 Hz. What is the energy associated with this frequency?

stellarik [79]

Use Planck's equation (E=hv) to solve. where frequency (v) of ultrviolet radiation is 6.8 × 1015 1/s. The variable h is a constant equal to 6.63 × 10-34 J·s
E= (6.8 × 1015 1/s)x(6.63 × 10-34 J·s)

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4 years ago

In the image above, the region labeled “ A “ is called the . . .

tatyana61 [14]

The awnser is central canal the third one

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2 years ago

Read 2 more answers

Having been heated to 800 K , at some point the tank starts to leak. By the time the leak is repaired, the tank contains only ha

Liula [17]

Answer: The temperature of the gas reduced to 400K.

Explanation:

Stated that ; The pressure remains the same, that is initial and final pressure equals 1atm.

Applying Charles Law

$V1/T1 = V2/T2$

Initial volume V1 = 1

Final volume V2 = 1/2 (halved)

Initial temperature T1 =800K

Final temperature T2 = ?

(1/800) = (1/2)/T2

T2 = 800/2

T= 400K

Therefore, when the volume is halved, the temperature reduced also to half ( 400K)

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3 years ago

In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m

salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : $Q=I\times t$

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec (1 min = 60 sec)

Now put all the given values in this formula, we get

$Q=10A\times 600s=6000C$

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = $2\times (1.6\times 10^{-19})$

Number of atoms deposited = $\frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22}$ atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = $\frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113$ moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g

Mass of Copper Deposited = $63.5\times 0.03113 =1.98g$

Therefore, the mass of copper deposit is, 1.98 grams

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3 years ago