How many molecules are there in 2.30 grams of NH3 ?

The following substituents are listed in random order. Using the Cahn–Ingold–Prelog convention, rank the groups below from highe

Ivahew [28]

Answer:

-SH, -CH2SH, - CH2CH2SH, -H,

7 0

3 years ago

Lridium-192 is an isotope of iridium and has a half-life of 73.83 days. if a laboratory experiment begins with 100 grams of irid

fiasKO [112]

Radioactive material undergoes first order dissociation kinetics.

For 1st order system,
k = 0.693 / t1/2
where, t 1/2 = half-life of the radioactive disintegration process.

Given that, t 1/2 = 73.83 days
Therefore, k = 0.009386 day-1

Also, for 1st order reaction,
k = $\frac{2.303}{t} log \frac{Co}{Ct}$

Given that, Co = initial concentration of Iridium-192 = 100 g

Therefore, 0.009386 = $\frac{2.303}{t} log \frac{100}{Ct}$

On rearranging we get, Ct = 100 $(0.990656)^{t}$

Answer: Ct = 100 $(0.990656)^{t}$ equation approximates the amount of Iridium-192 present after t days

8 0

3 years ago

What’s the difference between an ester and an amide

Rufina [12.5K]

In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. ... Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.

8 0

3 years ago

In order to become stable in an ionic bond, oxygen needs to (A) gain 2 electrons (C) share 2 electrons B) lose 2 electrons

Sedbober [7]

Answer:

gains 2 electrons. In order for the new ionic compound to form a a neutral compound, the oxygens must gain 6 electrons

3 0

3 years ago

Read 2 more answers

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

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 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$