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stealth61 [152]
2 years ago
6

How many moles of mercury 2 oxide are needed to produce 125 grams of oxygen

Chemistry
1 answer:
baherus [9]2 years ago
4 0
2 HgO >> 2 Hg + O2
moles O2 = 125 g / 32 g/mol =3.91
the ratio between HgO and O2 is 2:1
moles HgO needed = 2 x 3.91 =7.82
Hope this helps!
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If Maria winks exactly 5 times every minute while she is awake and she sleeps exactly 8 hours a day, how many times does Maria w
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24 minus 8 is 16
5 times 60 is 300
300 times 16 is 4800.
She winks 4800 times a day
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2 years ago
Titanium metal requires a photon with a minimum energy of 6.94×10⁻¹⁹J to emit electrons. What is the wavelength of this light? E
Elena-2011 [213]

Answer:2.86x10^-7m

Explanation:E=hc/^

E=6.94x10^-19J

c = 2.9979x10^8m/s

h= 6.626x10^-34Js

^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19

= 2.86x10^-7m

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2 years ago
Use the information in the square to answer the questions about copper. A purple box has C u at the center and 29 above. Below i
scoray [572]

Answer:

29 protons 29 electrons

34 neutrons or (65-29)=36 neutrons in its nucleus.

6 0
2 years ago
Read 2 more answers
A chemist wants to extract copper metal from copper chloride solution. The chemist places 1.50 grams of aluminum foil in a solut
Lisa [10]

Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

Explanation: 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.

Thus 54 g of of aluminium react with 270 g of copper chloride.

1.50 g of aluminium react with=\frac{270}{54}\times 1.50=7.5 gof copper chloride.

3 moles of copper chloride gives 3 moles of copper.

7.5 g of copper chloride gives 7.5 g of copper.

8 0
2 years ago
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What is the molarity of a solution that contains 122g of MgSO4 n 3.5L of solution?​
tatiyna

Answer:

0.29mol/L or 0.29moldm⁻³

Explanation:

Given parameters:

Mass of MgSO₄ = 122g

Volume of solution = 3.5L

Molarity is simply the concentration of substances in a solution.

Molarity = number of moles/ Volume

>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.

Number of moles = mass/ molar mass

Molar mass of MgSO₄:

Atomic masses: Mg = 24g

S = 32g

O = 16g

Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol

= (24 + 32 + 64)g/mol

= 120g/mol

Number of moles = 122/120 = 1.02mol

>>>> From the given number of moles we can evaluate the Molarity using this equation:

Molarity = number of moles/ Volume

Molarity of MgSO₄ = 1.02mol/3.5L

= 0.29mol/L

IL = 1dm³

The Molarity of MgSO₄ = 0.29moldm⁻³

6 0
2 years ago
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