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3 years ago

Give me lil reasoning so I know your not lying for points

Chemistry

1 answer:

Natali5045456 [20]3 years ago

4 0

Answer:

C. Tip had the lower pressure, it was 652.71mmHg

Explanation:

To answer this question you need to convert both values into a common unit so you can compare directly.

87kPa = 652.71 mmHg

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A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

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 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

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 = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

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

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

Attempt 1

Vsevolod [243]

Answer:

12.01

Explanation:

(12.00*98.93% + 13*1.07%) /100% = 12.01

3 0

3 years ago

10: Convert 77.0 L at 18.0 mmHg to its new volume at standard pressure.

olchik [2.2K]

Answer:

The answer to your question is V2 = 1.82 l

Explanation:

Data

Volume 1 = 77 l

Pressure 1 = 18 mmHg

Volume 2 = ?

Pressure 2 = 760 mmHg

Process

Use Boyle's law to solve this problem

P1V1 = P2V2

-Solve for V2

V2 = P1V1/P2

-Substitution

V2 = (18 x 77) / 760

-Simplification

V2 = 1386 / 760

-Result

V2 = 1.82 l

5 0

3 years ago

Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration

elixir [45]

Answer:

8.54

Explanation:

At equivalence point :

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :

pH = 1/2 [ pKw + pKa + log c ]

Ka for acetic acid = 1.75 X 10⁻⁵

so pKa = -log (1.75 X 10⁻⁵) = 4.74

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169

so log c = log 0.02169 = -1.66

putting the values....

pH = 1/2 [14 + 4.74 - 1.66 ]

pH = 1/2 [ 17.08] = 8.54

6 0

3 years ago

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Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would eff

kenny6666 [7]

Answer:

I was having trouble with this question idk the answer

Explanation:

4 0

3 years ago

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