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chubhunter [2.5K]
3 years ago
8

What is the volume of a balloon if it contains 3.2 moles of helium at a temperature of 20. °C and standard pressure?

Chemistry
1 answer:
KatRina [158]3 years ago
3 0

Answer:

1.14 mol

Explanation:

You might be interested in
The compound sodium thiosulfate pentahydrate, Na2S2O3 • 5H2O
Mnenie [13.5K]

The theoretical yield is 204.4 g while the percent yield is 2.57%.

<h3>What is theoretical yield?</h3>

Theoretical yield is the amount of product obtained based on the stoichiometry of the reaction.

S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)

Number of moles of sulfur =  3.25 g /8(32) = 0.013 moles

Number of moles of sodium sulfite =  13.1 g/126 g/mol = 0.103 moles

Since 1 moles of sulfur reacts with 8 moles of sodium sulfite

0.013 moles reacts with 0.013 moles ×  8 moles /1 mole = 0.104 moles

There is not enough sodium sulfite hence it is the limiting reactant.

1 mole of sodium sulfite yields 8 moles of product

0.103 moles of sodium sulfite yields  0.103 moles × 8 moles /1 mole = 0.824 moles

Mass of product = 0.824 moles × 248 g/mol = 204.4 g

percent yield =  5.26 g /204.4 g × 100/1

= 2.57%

Learn more about percent yield: brainly.com/question/2506978

4 0
2 years ago
By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling
vova2212 [387]

It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.

By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then

p_{1} =patm+pH_{2} O

p_{1} = 101325+ρgh_{1}

It is given that height is 125ft. Put the value of h in above formula:

h1  =125ft=38.1m

ρ=1.04g/mL=1040kg/m^{3}

g=9.81                                  

p_{1} =101325Pa+388711.44

p_{1}   =490036.44‬Pa                

p_{2} =p atm  =101325Pa

It is known that volume and pressure can be expressed as:

V*P=const.

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/101325

V_{2} /V_{1}=4.84

Assume constant temperature

d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.

now p_{1} =p atm​+pH_{2} O =490036.44Pa

V*p=const                              

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/X

p_{2} =490036.44pa/(V2/V1) =326690.96Pa

p_{2} =patm +pH_{2} O

p_{2} =101325Pa+ρgh_{2}

326690.96Pa=101325Pa+ρgh_{2}

ρgh1  =151987.5-101325=225365.96‬‬Pa

ρ=1,04g/mL=1040kg/m3

g=9.81

h_{2} =225365.96/‬ρ∗g

​h_{2}  =225365.96  / ‬1040∗9.81

h_{2}  =22.09m= 72.47ft

ΔH=H_{1} -H_{2}

=125-72.47

=52.53ft

So she can safely ascend up to 52.53 ft without Breathing out

To know more about Scuba diver here

brainly.com/question/15430942

#SPJ4

6 0
1 year ago
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

8 0
3 years ago
Can Hydrocyanic acid be a Lewis Acid? What about a Lewis Base?
FinnZ [79.3K]

HCN is a Bronsted acid; it can dissociate into H+ and CN-. And H+ is a Lewis acid because it accepts election pairs. ... In order for H+ and CN- to be formed, Hydrogen in HCN donates its electrons to Carbon. So in this sense, Hydrogen is the lewis base and Carbon is the lewis acid.

8 0
3 years ago
What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent
dsp73

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

         \text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, \text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

8 0
3 years ago
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