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3 years ago

HELP ME pls it’s in the picture below

Chemistry

2 answers:

sattari [20]3 years ago

7 0

Answer:

the IUPAC name is chromium(III) phosphate pentahydrate.

Ludmilka [50]3 years ago

6 0

Answer:

Hydrated chromium(III) phosphate or chromium(III) phosphate tetrahydrate.

You might be interested in

How many moles of carbon do you have in a trillion carbon

Dmitry_Shevchenko [17]

Answer:

$1.7\times 10^{-12}$ moles of carbon

Explanation:

1 trillion = $10^{12}$

Avogadro number :It is the number of particles that are present in 1 mole of substance.

$N_{0} = 6.022\times 10^{23}$

So,

$6.022\times 10^{23}$ atoms = 1 mole carbon

1 atom =

$\frac{1}{6.022\times 10^{23}}$ mole

1 trillion atom =

$\frac{1}{6.022\times 10^{23}}\times 10^{12}$ mole

$1.660\times 10^{-12}$ mol

Two significant figure

$1.7\times 10^{-12}$ moles of carbon

4 0

3 years ago

Please help fast, I will give brainliest.

frez [133]

Answer:

The reaction is favorable at all temperatures

Explanation:

Since G = H - TS, -H and +S would result in G = -H -TS, which will always be negative.

7 0

3 years ago

3. A PS6 plugged into a 120 V outlet uses 3 Amps. What is the Resistance rating of

blondinia [14]

Answer:

R = 40 ohms

Explanation:

Given that,

The voltage of outlet, V = 120 V

Current flowing through the device, I = 3 A

We need to find the Resistance rating of the power cord. Let the resistance is R. We know that,

Ohm's law, V = IR

Put all the values,

$R=\dfrac{V}{I}\\\\R=\dfrac{120}{3}\\\\R=40\ \Omega$

So, the resistance rating of the power cord is equal to 40 ohms.

8 0

3 years ago

Which half-reaction can occur at the anode in a

alexira [117]

The answer is (3). The reaction that can occur at the anode is oxidation reaction which will lose electrons. So (1) and (2) are not correct. For (4) Fe3+ can not lose electrons again.

4 0

4 years ago

If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em

erica [24]

Answer: Wavelength associated with the fifth line is 397 nm

Explanation:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $10973731.6m^{-1}$

$n_f$ = Higher energy level = 7

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

$\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2$

$\frac{1}{\lambda}=2.52\times 10^{6}m$

$\lambda}=3.97\times 10^{-7}m=397 nm$ $1nm=10^{-9}m$

Thus wavelength λ associated with the fifth line is 397 nm

7 0

4 years ago