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3 years ago

How can so many different substances in the world be made of so few elements?

1 answer:

7 0

Matematically speaking, maybe because:
The number of substances = number of elements + number of different combinations of those elements

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(g) The mole fraction of O2 in atmospheric air is 0.21. Calculate the concentration of O2 in a sample of atmospheric air in a 10

irinina [24]

Answer:

The concentration of O₂ in the sample is 5,1x10⁻³ M (mol/L)

Explanation:

To obtain the moles of gas you should use ideal gas formula:

PV/RT = n

Where:

P is pressure: 650 mmHg or 650 mmHg* ( 1 atm / 760 mmHg) = 0,855 atm

V is volume: 10 L

R is gas constant: 0,082 atm·L/mol·K

T is temperature: 160°C or 160 + 273,15 = 433,15 K

And n is mol number

Thus, replacing bold values in ideal gas formula:

n = 0,24 mol

The mole fraction of O₂ in atmospheric air is 0,21. Thus, the moles of O₂ there are:

0,24 moles of atmospheric air ₓ 0,21 = 0,051 O₂ moles

There are many ways to express concentration but we will express it in molarity (mol of solute / L of solution) understanding solute as O₂ and solution as atmospheric air. So:

0,051 O₂ moles / 10 L = 5,1x10⁻³ M -<em>M is molarity</em>-

I hope it helps!

4 0

3 years ago

Help please, 44 to 45, calculate the answers to the following problems. Use the following equation as the basis of your calculat

wlad13 [49]

Answer: 44,8 l. of CO2 and 72 g. of water will be produced

Explanation:

6 0

3 years ago

Read 2 more answers

Gases present in atmosphere are in ionic form true or false

ANTONII [103]

Answer:

false

Explanation:

Only ionic compounds can dissolate in water.

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

How does a buffer resist change in ph upon addition of a strong acid?

lora16 [44]

Any buffer exists in this equilibrium

HA <=> $H^+ + A^-$
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base ( $A^-$ )

When a strong acid is added, it reacts with the large reservoir of the conjugate base ( $A^-$ ) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.

8 0

3 years ago