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3 years ago

Can someone help?? This is really hard.

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

8 0

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4 /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

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Draw a Lewis structure for SO 2 in which all atoms obey the octet rule. Show formal charges. Do not consider ringed structures.

blagie [28]

Answer:

The answer is attached below

Explanation:

To draw the Lewis structure the sulphur willl be placed in the centre with the Valence electron sorrounding it the electrons between sulphu and oxygen to form bonds

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3 years ago

1 mole solution of NaCl contains

Mamont248 [21]

Answer:

If you dissolve 58.44g of NaCl in a final volume of 1 liter, you have made a 1M NaCl solution, a 1 molar solution.

Explanation:

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3 years ago

determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass

Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

$K_{f}$ or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

Δ $T_{f}$ = i × $K_{f}$ ×m

T⁰- $T_{s}$ = $i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water in kg}$

in which,

Δ $T_{f}$ = change in freezing point

Δ $T_{s}$ = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T $_{s}$ = 1 ×1.86°C/m × $\frac{30.7}{92.09g/mol * 0.376kg}$

$T_{s}$ = -1.65°C

Thus, the freezing point depression of a solution is -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

Learn more about freezing point depression :

brainly.com/question/26525184

#SPJ4

8 0

1 year ago

In an experiment, the molar mass of the compound was determined to be 118.084 g/mol. What is the molecular formula of the compou

pishuonlain [190]

Answer:

Succinic acid

Explanation:

The most common possibility is succinic acid

As it has decimals after whole no till hundredth it contains OH and C in most of the cases .

Let's check for succinic acid

C4H_6O_6
4(12)+4(16)+6
64+48+6
118u

Yes approximately equal

Molecular formula is.

(CH_2)_2(CO_2H)_2

3 0

2 years ago

How many moles of Carbon are in 3.06 g of Carbon

natta225 [31]

Answer:

$\boxed {\boxed {\sf 0.255 \ mol \ C }}$

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

$\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Since we are converting 3.06 grams to moles, we multiply by that value.

$3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}$

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

$3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}$

$3.06 * \frac{1 \ mol \ C}{12.011 }$

$\frac {3.06}{12.011 } \ mol \ C$

$0.25476646 \ mol \ C$

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

$0.255 \ mol \ C$

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>

3 0

2 years ago