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mihalych1998 [28]
2 years ago
11

In order to wash out the impurity, you looked up some sketchy method from 1952 which Hank Pym wrote during his Ph. D. under E. J

. Corey. The method requires you add HCl to the mixture. After you rid your sample of any impurities, you take some pH paper and find out your solution has a pH of 3.0. Assuming the pKa of Pym Particles is 4.5 and that the molar absorptivity of the protonated Pym Particle is 4.82 * 102 L/mol*cm and 8.54 * 103 L/mol*cm for 400 nm and 690 nm respectively, calculate the expected A400 and A690 of the solution.

Chemistry
1 answer:
Eva8 [605]2 years ago
8 0

Answer:

516.24

Explanation:

The solved solution is in the attached document.

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What mass of water (in grams) is produced by the reaction of 23.0 g of SiO2?
nikitadnepr [17]

The mass of water produced by the reaction of the 23 g of SiO_2  is 13.8 g.

The given chemical reaction;

4Hf (g)  \ + \ SiO_2 (s) \ --> \ SiF_4(g) \ + \ 2H_2O(l)

In the given compound above, we can deduce the following;

  • molecular mass of SiO_2 = 28 + (2 x 16) = 60 g
  • molecular mass of 2H_2O = 2(18) = 36 g

60 g of SiO_2  --------- 36 g of water

23 g of SiO_2  ------------- ? of water

mass \ of \ water = \frac{23 \times 36}{60} = 13.8 \ g \ of \ water

Thus, the mass of water produced by the reaction of the 23 g of SiO_2  is 13.8 g.

  • <em>"Your question is not complete, it seems to be missing the following information";</em>

In the reaction of the given compound, 4Hf (g)  \ + \ SiO_2 (s) \ --> \ SiF_4(g) \ + \ 2H_2O(l), what mass of water (in grams) is produced by the reaction of 23.0 g of SiO2?

Learn more here:brainly.com/question/13644576

5 0
2 years ago
Brain if correct<br><br><br> (3)
Anni [7]

Answer:

Darker colored items absorb more of the sunlight.

Explanation:

Lighter colored item tend to reflect more of the sunlight

8 0
2 years ago
Read 2 more answers
Why can scratching at the bottom of a flask induce crystallization?
Ksenya-84 [330]

Scratching causes cracks and crevices on the surface of the flask (though microscopically). These will act as favorable sites for nucleation, which leads to the formation of crystals.

7 0
2 years ago
AGO H2SO4 (aq) Al(SO4)3(aq) H28) fe Cl218) FeCl38<br>​
Rina8888 [55]

Answer:

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

2Fe + 3Cl2 → 2FeCl3

Explanation:

1. (SO4) 3 you see this 3 it means that 3 must be behind H2SO4. So now it's 3H2SO4.

2. If 3 is now behind one H2, it must be behind the other.

So now it's 3H2.

3. Al2 (SO4) 3 has 2 ahead of Al which means there will be 2Al in the reactants.

1. FeCl3 has 3 ahead of Cl, and Cl2 has 2. Which means that behind FeCl3 goes 2, and behind Cl2 goes 3 so now we have equated all Cl.

2. Since it is now 2FeCl3, we know that there must be 2 in the second Fe. It's 2Fe now.

7 0
2 years ago
A 10.000g ice cube is added to 100.000g of water at 80.0 oC in a calorimeter. The final temperature of the water in the calorime
Ket [755]

<u>Answer:</u> The amount of heat needed to melt the ice cube is 6.083 kJ

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(65.5^oC,338.5K)

  • <u>For process 1:</u>

To calculate the amount of heat required to melt the ice at its melting point, we use the equation:

q_1=m\times \Delta H_{fusion}

where,

q_1 = amount of heat absorbed = ?

m = mass of ice = 10.000 g

\Delta H_{fusion} = enthalpy change for fusion = 334.16 J/g

Putting all the values in above equation, we get:

q_1=10.000g\times 334.16J/g=3341.6J

  • <u>For process 2:</u>

To calculate the heat required at different temperature, we use the equation:

q_2=mc\Delta T

where,

q_2 = heat absorbed

m = mass of ice = 10.000 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=[65.0-0]^oC=65.5^oC

Putting values in above equation, we get:

q_2=10.000g\times 4.184J/g.^oC\times 65.5^oC\\\\q_2=2741.83J

Total heat absorbed = q_1+q_2

Total heat absorbed = [3341.6+2741.83]J=6083.43J=6.083kJ

Hence, the amount of heat needed to melt the ice cube is 6.083 kJ

4 0
3 years ago
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