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harina [27]
3 years ago
9

How many moles are in 395 grams of KMnO4 if the molar mass is 158 g/mole? please help me

Chemistry
1 answer:
Gala2k [10]3 years ago
8 0

Answer:

I have no idea

Explanation:

I don’t know

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Consider the Bohr model of the simplest atom, hydrogen. What energy photon will be released if an electron falls from the n= 2 o
Nana76 [90]
<h2>Ultraviolet Light</h2>

Explanation:

  • The energy of a photon that will be released if an electron falls from the n= 2 orbit (excited state) to the n0 = 1 orbit (ground state) is of ultraviolet light.
  • In the ultraviolet part of the spectrum, a photon having an energy of 10.2 eV has a wavelength of 1.21 x 10-7 m.
  • Hence, when an electron wants to jump or it gets excited from the first level to the second level that is from n = 1 orbit to n = 2 orbits, it must absorb a photon of ultraviolet light.
  • But,When an electron falls from n = 2 orbit to n = 1 orbit  or from n = 2 orbit(excited state) to n = 0 orbit(groubd state), it emits a photon of ultraviolet light.
7 0
3 years ago
A person’s genotype is pp. p=five fingers. What is this person’s phenotype?
Kay [80]

Answer:

25

Explanation:

5×5 =25

3 0
3 years ago
How many electrons are in the highest occupied energy level of a neutral chlorine atom?
tino4ka555 [31]
There are seven electrons
5 0
3 years ago
Read 2 more answers
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
BigorU [14]
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).

The first step is to find the number of moles of Mg in 4.03g of Mg.  You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg.  Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg.  To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂.  From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP.  Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.
6 0
3 years ago
Read 2 more answers
The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were a
melomori [17]

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

                          (100ml - 1.468 ml)

                        = 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

          Amount of water = \frac{1.468 ml \times 23 ml}{98.532 ml}

                                       = 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

                Mg + H_{2}O \rightarrow Mg(OH)_{2}

Molar mass of Mg = 24.305 g

Molar mass of H_{2}O = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

           Amount of Mg = \frac{24.305 g \times 0.3427 ml}{18 g}  

                                    = 0.462 g

   

6 0
3 years ago
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