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valentinak56 [21]
4 years ago
12

At constant temperature and pressure, 5.0 L of SO2 is

Chemistry
1 answer:
Ne4ueva [31]4 years ago
4 0
I've done this a few times and keep coming up with 5.0 L . I used the mole ratio, and SO 2 as the LR. That would make 2 moles of SO3 5.0 L but its not one of your choices. the temp and pressure are constant, so according to n/v = p/rt the volume is the same as the moles. hope this helps
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Which of these equations is balanced? a. 3 H2SO4 + 2Al → Al2(SO4)3 + 3H2 b. 2KCl + Pb(NO3)2 → KNO3 + PbCl2 a b a and b neither
wolverine [178]
A.) 3 H2SO4 + 2 Al > Al2(SO4)3 + 3 H2
6 H+ +3 S +12 O +2 Al> 2 Al + 3S + 12O + 6H+
This one is balanced, with 6 H, 3 S, 12 O, and 2 Al on both sides.
B.) 2KCl + Pb(NO3)2 > KNO3 + PbCl2
2K + 2Cl + Pb + 2N +6 O > 1 K + 1 N+ 3 O + Pb + 2 Cl
This one is not balanced as there is 2 K, 6 O and 2 N on the reactants side but only 1 K, 3 O, and 1 N on the products side.
Therefore, only A.) is balanced
8 0
3 years ago
Read 2 more answers
1.
NeTakaya

1.     5

5 electrons

Hydrogen shares its only electron with Carbon to get a full valence shell. Now Carbon has 5 electrons. Because each Carbon atom has 5 electrons--1 single bond and 3 unpaired electrons--the two Carbons can share their unpaired electrons, forming a triple bond

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5 0
3 years ago
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Can colloidal suspensions be separated out by filtration
balu736 [363]

Answer:

Colloidal can not be separated through filtration.

Suspension can be separated through filtration.

Explanation:

Colloidal:

Colloidal consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm. While the pore size of filter paper is 2μm. That's why we can not separate the colloidal through the filtration. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.

Suspension:

The particle size in suspension is greater than 1000 nm. The particles in suspension can be separated through the filtration. These particles are large enough and can be seen through naked eye.

5 0
3 years ago
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PLEASEE HELPPP ME WITH NUMBER 5!!!!!!!
riadik2000 [5.3K]

Answer:

hi dont report me we are friends can i get brainliest plzz?

Explanation:

5 0
3 years ago
Radioactive cobalt-60 is frequently used in treating cancer. It took 24 years for a 10 gram sample to decay to 0.625 grams. What
-BARSIC- [3]

Answer:

Half life is 6 years.

Explanation:

T½ = In2 / λ

Where λ = decay constant.

But N = No * e^-λt

Where N = final mass after a certain period of time

No = initial mass

T = time

N = 0.625g

No = 10g

t = 24 years

N = No* e^-λt

N / No = e^-λt

λ = -( 1 / t) In N / No (inverse of e is In. Check logarithmic rules)

λ = -(1 / 24) * In (0.625/10)

λ = -0.04167 * In(0.0625)

λ = -0.04167 * (-2.77)

λ = 0.1154

T½ = In2 / λ

T½ = 0.693 / 0.1154

T½ = 6.00 years.

The half life of radioactive cobalt-60 is 6 years

5 0
3 years ago
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