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Firlakuza [10]
3 years ago
14

Question 10

Chemistry
1 answer:
zepelin [54]3 years ago
3 0
Because the ring is hollow
You might be interested in
Which statement below describes the behavior of a Molecules when a substance changes from a gas to liquid
krek1111 [17]

Answer:

As a substance changes from a solid to a liquid to a gas, its molecules first the molecules are moving fast enough, they are able to "escape." They leave the surface of the liquid as gas molecules. Evaporation is not the only process that can change a substance from a liquid to a gas. The same change can occur through boiling.

Explanation:

hopfully this helps!

7 0
3 years ago
Why is the speed of these particles deffirent in each state
monitta

Answer:

due to difference ic kinetic energy

Explanation:

3 0
3 years ago
If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

8 0
3 years ago
CsBr formula name???
olganol [36]

Answer

PubChem CID/molecular formula

Explanation:

Cesium bromide

PubChem CID 24592

Molecular Formula CsBr or BrCs

Synonyms CESIUM BROMIDE 7787-69-1 Caesium bromide Cesiumbromide Cesium bromide (CsBr) More...

Molecular Weight 212.81 g/mol

Component Compounds CID 260 (Hydrogen bromide) CID 5354618 (Cesium)

have a good day /night

may i please have a branllist

4 0
3 years ago
If the temperature on 244 mL of a gas is changed to 488 mL and 6 atm, at constant
Fittoniya [83]

Answer:

<h2>12 atm</h2>

Explanation:

To find the initial pressure we use the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the initial pressure

P_1 =  \frac{P_2V_2}{V_1}  \\

From the question we have

P_1 =  \frac{488 \times 6}{244}  =  \frac{2928}{244}  \\

We have the final answer as

<h3>12 atm</h3>

Hope this helps you

6 0
3 years ago
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