Answer:
It lies in in the <u>visible</u><u> </u><u>region</u><u>.</u>
Frequency: <u>is</u><u> </u><u>7</u><u>.</u><u>4</u><u>0</u><u>7</u><u> </u><u>×</u><u> </u><u>1</u><u>0</u><u>^</u><u>-</u><u>1</u><u>4</u><u> </u><u>Hz</u>

Explanation:
V is speed of light.
f is frequency
lambda is the wavelength
add all the number and find the average then subtract the mass defect and then you will get your answer
Answer:
Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.
Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.
Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.
Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.
Re-weigh the crucible and contents once cold.
Calculation:
Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)
Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment
Calculate the number of moles of anhydrous copper(II) sulfate formed
Calculate the number of moles of water driven off
Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed
Write down the formula for hydrated copper(II) sulfate.
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Explanation:
Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+