What are the empirical formula and empirical formula mass for N2S3

You have a solution that is 0.02M formate (HCOO-) and 0.03M formic acid (HCOOH), which has a Ka of 1.8x10-4. What is the pH of t

lisov135 [29]

Answer:

3.6

Explanation:

Step 1: Given data

Concentration of formic acid: 0.03 M

Concentration of formate ion: 0.02 M

Acid dissociation constant (Ka): 1.8 × 10⁻⁴

Step 2: Calculate the pH

We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.

$pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6$

6 0

3 years ago

Heating curve shows temperature verses energy gain. Which parts of the curve represent a gain in potential energy?

Brilliant_brown [7]

Answer:

Those two horizontal lines.

Explanation:

Hello there!

In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.

In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.

Regards!

6 0

2 years ago

A compound is composed of 13% carbon, 4.3% hydrogen, 30.4% nitrogen, and 52.2% oxygen. The mystery compound has a molar mass of

vredina [299]

Answer:

C₂H₈N₄O₆ is the molecular formula for the compound

Explanation:

Data from the problem:

13 g of C in 100 g of compound

4.3 g of H in 100 g of compound

30.4 g of N in 100 g of compound

52.2 g of O in 100 g of compound

Firstly we determine, the mass of each in 184 g of compound, which is 1 mol

(13 g / 100 g) . 184 g = 24 g C

(4.3 g / 100 g) . 184 g = 7.91 g H ≅ 8 g H

(30.4 g / 100 g) . 184 g = 56 g N

(52.2 g / 100 g) . 184 g = 96 g O

And now, we divide the mass by the molar mass of each to determine the moles:

24 g C / 12 g/mol = 2 mol C

8g H / 1 g/mol = 8 mol H

56 g N / 14 g/mol = 4 mol N

96 g O / 16 g/mol = 6 mol O

So the molecular formula of the compound is C₂H₈N₄O₆

8 0

3 years ago

What part of the electromagnetic spectrum was first developed during ww2

insens350 [35]

Answer:

Radio waves were first developed in WWII

Explanation:

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7 0

3 years ago

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

Andru [333]

Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

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 $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

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 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

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

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

6 0

2 years ago