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dsp73
3 years ago
10

If you were asked to covert 1.3 mol al to atoms which of the following should you use for the concersion

Chemistry
1 answer:
Gnoma [55]3 years ago
4 0
1 mole/ (6.022x10^23)
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How many of the following molecules are polar? Pcl5 cos xeo3 sebr2
klasskru [66]

Answer:

<em>Three (3) of the molecules are polar: </em>CoS,<em> </em>XeO_{3},<em> </em>SeBr_{2}<em>.</em>

Explanation:

Polar substances have their elements held together by a covalent bond that contain partially positive and negative charges, which results in a difference in the charges' electronegativity difference (usually ranging between 0.4 and 0.7).

  • PCl5 is <u>non-polar</u> with a symmetric geometry
  • CoS is <u>polar</u>
  • XeO3 is <u>polar</u>, with a trigonal pyramidal molecular geometric
  • SeBr2 is <u>polar</u> as the difference their electronegativity is about ).5
5 0
3 years ago
The difference between diffusion and osmosis is that osmosis deals only with ________.
Nat2105 [25]
Osmosis deals only with D. Water. Diffusion and Osmosis are relatively the same thing besides the fact that water is largely incorporated with the osmosis.
6 0
3 years ago
The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B
loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

C_{15}H_{24}0

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in CO_2 and also, all the hydrogen in BHT is  present in H_2O, Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of  CO_2 = 8.990 g*(\frac{1mole}{44.01g})

                       =  0.2043 moles

∴ moles of C =  0.2043 moles

moles of H_2O = 2.944 g *(\frac{1mole}{18.01g} )

                       = 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

                      = 0.327 moles

Since number of moles= \frac{mass}{molarmass}

number of moles of H =  0.327 moles

molar mass of H = 1.008 g/mol

∴  mass of H in the sample = 0.327 moles × 1.008 g/mol

                                             = 0.329616g

                                             

mass of C in the sample can be calculated as = 0.2043 moles × (\frac{12.01g}{1 mole} )

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × (\frac{1mole}{16.0g})

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C                                          H                                    O

0.2043                                0.327                             0.0136

Dividing by the least number (0.0136) , we have :

\frac{0.2043}{0.0136}                                     \frac{0.327}{0.0136}                               \frac{0.0136}{0.0136}

15.02                                      24.04                             1

≅

15                                             24                                  1

Therefore, the empirical formula would be : C_{15}H_{24}0

7 0
3 years ago
What is the name of CaCO2
weqwewe [10]
Calcium--Cobalt(1/2)
8 0
3 years ago
Heeeeellllllppppppp!!!!!
ycow [4]

Answer:

Heterogeneous

Explanation:

3 0
2 years ago
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