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Ostrovityanka [42]
3 years ago
8

NEED ASAP

Chemistry
1 answer:
notka56 [123]3 years ago
5 0

Answer: B

Refer:

Because water molecules are polar, they interact with the sodium and chloride ions. In general, polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes. This concept is often expressed as “Like dissolves like.

You might be interested in
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
If 35.0 mL of water in a graduated cylinder is displaced by 8.00 mL
sdas [7]

Answer:

<h2>0.52 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}\\

From the question

volume = final volume of water - initial volume of water

volume = 35 - 8 = 27 mL

We have

density =  \frac{14}{27}  \\  = 0.518518...

We have the final answer as

<h3>0.52 g/mL</h3>

Hope this helps you

3 0
2 years ago
_Al + _Fe(NO3)2 --&gt;_Fe + __Al(NO3)3
romanna [79]
4Al+6Fe(NO3)2= 6Fe+4Al(NO3)3
4 0
3 years ago
PLEASE HELP
swat32

Answer:

1461.7 g of AgI

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

5 0
3 years ago
Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a
svetoff [14.1K]

Answer:

The answer to your question is        P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

                      = 373°K

Formula

               (P + \frac{a}{v^{2}} )(v - b) = RT

Substitution

               (P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)

Simplify

               (P + 0.0094)(22.3829) = 30.586

Solve for P

                           P + 0.0094 = \frac{30.586}{22.3829}

                           P + 0.0094 = 1.366

                                 P = 1.336 - 0.0094

                                P = 1.357 atm

7 0
3 years ago
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