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Aloiza
2 days ago
15

What is the molar mass of citric acid (C6H8O7) and baking soda (NaHCO3)?

Chemistry
1 answer:
Alekssandra [968]2 days ago
4 0

Answer:

1. 192.0 g/mol.

2. 84.0 g/mol.

Explanation:

  • The molar mass refers to the weight of all atoms combined in a molecule measured in grams per mole.
  • To find a molecule's molar mass, we begin by looking up the atomic weights of the relevant elements from the periodic table. Next, we tally the atoms present and multiply that by their respective atomic weights.

1. Molar mass of citric acid (C₆H₈O₇):

Molar mass of C₆H₈O₇ = 6(atomic mass of C) + 8(atomic mass of H) + 7(atomic mass of O) = 6(12.0 g/mol) + 8(1.0 g/mol) + 7(16.0 g/mol) = 192.0 g/mol.

2. Molar mass of baking soda (NaHCO₃):

Molar mass of NaHCO₃ = (atomic mass of Na) + (atomic mass of H) + (atomic mass of C) + 3(atomic mass of O) = (23.0 g/mol) + (1.0 g/mol) + (12.0 g/mol) + 3(16.0 g/mol) = 84.0 g/mol.

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Answer:

The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in  15.3 \frac{atoms}{min-g}

The dissolution rate of the sample is given by 9.16 \frac{atoms} {min-gram}

The C-14 proportion within the sample can be determined as per = \frac {9.16}{15.3} \\\\ = 0.5987

With a half-life of 5730 years.

Now, we need to compute the number of half-lives (n) that are applicable:

(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987\\\\

\therefore \\\\ \Rightarrow n= \frac{0.227}{0.3010} \\\\ = 0.740\\

Thus, the age of the sample is represented as = n \times\ half-life

                                                 = 0.740 \times 5730 \ years \\\\=4241.17 \ years\\\\

6 0
1 day ago
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u
lions [985]

Answer:

The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.

Explanation:

Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Thus, the overall reaction becomes;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?

According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;

Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

7 0
1 day ago
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [956]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

4 0
14 days ago
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Write a balanced equation for the reaction of NaCH3COO (also written as NaC2H3O2) and HCl.
lorasvet [956]
The balanced equation is:

NaCH₃COO + HCl → NaCl + HCH₃COO

Make 
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10 days ago
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At 1.01 bar, how many moles of CO2 are released by raising the temperature of 1 litre of water from 20∘C to 25∘C
VMariaS [1037]

Answer: 0.0007 moles of CO_2 are released when the temperature rises.

Explanation:

To determine the moles, we utilize the ideal gas law, as follows:

PV=nRT

where,

P = gas pressure = 1.01 bar

V = gas volume = 1L

R = gas constant = 0.08314\text{ L bar }mol^{-1}K^{-1}

  • Calculated moles at T = 20° C

The gas temperature = 20° C = (273 + 20)K = 293K

Substituting values into the equation gives:

1.01bar\times 1L=n_1\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 293K\\n_1=0.04146moles

  • Calculated moles at T = 25° C

The gas temperature = 25° C = (273 + 25)K = 298K

Substituting values into the equation gives:

1.01bar\times 1L=n_2\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 298K\\n_2=0.04076moles

  • Released moles = n_1-n_2=0.04146-0.04076=0.0007moles

Therefore, 0.0007 moles of CO_2 are released when the temperature increases from 20° C to 25° C.

5 0
13 days ago
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