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Pepsi [2]
3 years ago
10

The density of helium in a 2.00 L tank at 1.0 atm and 23 degrees Celsius is?

Chemistry
1 answer:
andrezito [222]3 years ago
5 0

The output density is given as kg/m 3, lb/ft 3, lb/gal(US liq) and sl/ft 3. Specific weight is given as N/m 3 and lb f / ft 3.

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Rank the ions in each set in order of decreasing size, and explain your ranking:
iragen [17]

Ranking of the ions in each set in order of decreasing size -

Cs+ > Ba²⁺ > Sr²⁺

Cs+ has the biggest size, because its more downward (compared to Sr2+) and more to the left (compared) ot Ba2+.

Sr2+ has the smallest size because it's more upwords (compared to Cs+ and Ba2+).

For isoelectronic ions, the greater the nuclear charge , the greater is the attraction for electrons and smaller is ionic radius. In other words, size of the isoelectronic ions decreases with increase in atomic number.

Cl3+ has the smallest ion among all.

To learn more about ions from given link

brainly.com/question/14658491

#SPJ4

5 0
2 years ago
Which temperature represents the highest average kinetic energy of the particles in a sample of matter ?
Genrish500 [490]

Answer: Option C is correct.

Explanation: Average kinetic energy is directly proportional to the absolute temperature. Higher the temperature means higher the kinetic energy.

Average kinetic energy is given by:

K.E._{avg}=\frac{3kT}{2}

Where, k = Boltzman constant

T = Temperature

We are given different temperatures, so to compare they all should have the same units.

a) 298K

b) 267K

c) 27°C = 273+27 = 300K

d) 12°C = 273+12 = 285K

Looking at the temperature values, C part will have the highest average kinetic energy.

3 0
3 years ago
Read 2 more answers
If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the mola
rodikova [14]
Volume Ba(OH)2 = 23.4 mL in liters : 

23.4 / 1000 => 0.0234 L

Molarity  Ba(OH)2 = 0.65 M

Volume HNO3 = 42.5 mL in liters:

42.5 / 1000 => 0.0425 L

number of moles Ba(OH)2 :

n = M x V

n = 0.65 x 0.0234 

n = 0.01521 moles of Ba(OH)2

Mole ratio :

<span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
</span>
1 mole Ba(OH)2 ---------------- 2 moles HNO3
 0.01521 moles ----------------- moles HNO3

moles HNO3 = 0.01521 x 2 / 1

moles HNO3 = 0.03042 / 1

= 0.03042 moles HNO3

Therefore:

M ( HNO3 ) = n / volume ( HNO3 )

M ( HNO3 ) =  0.03042 / 0.0425

M ( HNO3 ) = 0.715 M

5 0
3 years ago
Why does more carbon dioxide cause the water to be acidic?
joja [24]
<span>When carbon dioxide is passed through water, some of it dissolves. A small fraction of the dissolved CO2 interacts with the water to become carbonic acid, H2CO3.</span>
6 0
3 years ago
Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
pochemuha

Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

First we have to calculate the  mass of HCO_3^- in tablet.

\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

6 0
3 years ago
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