What is the mass in grams of 6.022×1023 atoms of mass 16.00 amu?

Henri becquerel and the curies found out that _____.

natulia [17]

Henri becquerel and the curies found out that atoms were not indivisible and indestructible

Explanation:

The French Physicist, Henri Becquerel, created the appearance of radioactivity. He shared the Nobel Prize in Physics with Marie Curie and Pierre Curie (Marie's husband) in 1903 for their work in radiation. A radionuclide will emit transmission through the process of radioactive sense.

5 0

2 years ago

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ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

Answer: The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

2 years ago

Room temperature is about 20 degrees Celsius. Explain how you could convert this temperature to kelvin. Use evidence from the fi

Taya2010 [7]

Answer:

293.15 K.

Explanation:

It is given that, the room temperature is 20 degrees Celsius.

We need to convert this temperature into kelvin.

The conversion from degrees Celsius to Kelvin is as follows :

$T_k=T_c+273.15$

We have, $T_c=20^{\circ} C$

So,

$T_k=20+273.15\\\\T_k=293.15\ K$

So, the room temperature is 293.15 kelvin.

8 0

2 years ago

(A) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pus

balandron [24]

Answer:

(A) The work done by the system is -101.325J

(B) The workdone by the system is -90.75J

Explanation:

(A) Workdone = -PΔV

Given that A = 100cm2 = 0.01m2

distance d = 10cm = 0.1m

ΔV= Area × distance

ΔV= 0.01 ×0.1

ΔV = 0.001m3

P= external pressure = 1atm = 101325Pa

Workdone = -0.001 × 101325

W= - 101.325Pa m3

1Pam3 = 1J

Therefore W = - 101.325J

The work done on the system is -101.325J

(B) Workdone = -PΔV

Given that A = 50cm2 = 0.005m2

distance d = 15cm = 0.15m

ΔV= Area × distance

ΔV= 0.005×0.15

ΔV = 0.00075m3

P=121kPa = 121000Pa

W= - 121000 × 0.00075

W= -90.75Pa m3

1Pam3 = 1J

W = - 90.75J

The woekdone by the system is -90.75J

5 0

2 years ago