CHEMISTRY HIGH SCHOOL

The percent composition by mass of certain compound is 35.5 percent carbon, 4.8 percent hydrogen, 37.9 percent oxygen, 8.3 percent nitrogen, and 13.5 percent sodium. What is the molecular formula if the molar mass is roughly 170 grams per mole

Answers

Answer 1

Answer:

Answer:

$C_(5)H_(8)NNaO_(4)$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 35.5

Molar mass of C = 12.0107 g/mol

% moles of C = $(35.5)/(12.0107)$ = 2.95569

% of H = 4.8

Molar mass of H = 1.00784 g/mol

% moles of H = $(4.8)/(1.00784)$ = 4.76266

% of N = 8.3

Molar mass of N = 14.0067 g/mol

% moles of N = $(8.3)/(14.0067)$ = 0.59257

% of Na = 13.5

Molar mass of Na = 22.989769 g/mol

% moles of N = $(13.5)/(22.989769)$ = 0.58721

% of O = 37.9

Molar mass of O = 15.999 g/mol

% moles of O = $(37.9)/(15.999)$ = 2.36889

Taking the simplest ratio for C, H, O, N, Na and O as:

2.95569 : 4.76266 : 0.59257 : 0.58721 : 2.36889

= 5 : 8 : 1 : 1 : 4

The empirical formula is = $C_(5)H_(8)NNaO_(4)$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12*5 + 1*8 + 14 + 23 + 16*4 = 169 g/mol

Molar mass = 170 g/mol

So,

Molecular mass = n × Empirical mass

170 = n × 169

⇒ n ≅ 1

The molecular formula = $C_(5)H_(8)NNaO_(4)$

Answer 2

Answer:

Answer : The molecular formula of a compound is, $C_5H_8O_4NNa$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 35.5 g

Mass of H = 4.8 g

Mass of O = 37.9 g

Mass of N = 8.3 g

Mass of Na = 13.5 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Molar mass of N = 14 g/mole

Molar mass of Na = 23 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (35.5g)/(12g/mole)=2.96moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (4.8g)/(1g/mole)=4.8moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (37.9g)/(16g/mole)=2.37moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (8.3g)/(14g/mole)=0.593moles$

Moles of Na = $\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= (13.5g)/(23g/mole)=0.587moles$

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

For C = $(2.96)/(0.587)=5.0\approx 5$

For H = $(4.8)/(0.587)=8.1\approx 8$

For O = $(2.37)/(0.587)=4.0\approx 4$

For N = $(0.593)/(0.587)=1.0\approx 1$

For Na = $(0.587)/(0.587)=1$

The ratio of C : H : O : N : Na = 5 : 8 : 4 : 1 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_5H_8O_4N_1Na_1=C_5H_8O_4NNa$

The empirical formula weight = 5(12) + 8(1) + 4(16) + 1(14) + 1(23) = 169 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=(170)/(169)=1$

Molecular formula = $(C_5H_8O_4NNa)_n=(C_5H_8O_4NNa)_1=C_5H_8O_4NNa$

Therefore, the molecular of the compound is, $C_5H_8O_4NNa$

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If the hydrogen ion concentration is 10-7M, what is the pH of the solution?Show all work.

Answers

Remember pH=-log(H ions). So it would be pH=-log(10^-7).

The specific heat of a solution is 4.18 J/(g•°C)and its density is 1.02 g/mL. The solution is formed by combining 25.0 mL of solution A with 25.0 mL of solution B with each solution initially at 21.4°C. The final temperature of the combined solution is 25.3°C. Calculate the heat of reaction, q, assuming no heat loss due to the calorimeter. I got 831 J
The part I cannot figure out is the question afterwards which is If the calorimeter has a heat capacity of 8.20 J/°C and a correction is included to account for the heat absorbed by the calorimeter what is the heat of reaction q

Answers

Answer:

a. qrxn = 831 J

b. 863 J

Explanation:

we know that density is the mass of a substance per unit volume

d=mass/volume

the volume of the solution is the combination of solution A and solution B

1.02 g/mL=mass/(25+25)

mass=50*1.02

mass=51g

Recall that Q=mCdT

mass=m, C=specific heat capacity

dT=change in temperature

qrxn = (51 g)(4.18 J/g⋅°C)(25.3 °C - 21.4 °C)

qrxn = 831 J

2.Heat=Heat capacity *change in temperature

qcal = (8.20 J/°C)((25.3 °C - 21.4 °C)

qcal = 31.98 J

qrxni + qcal = qrxn

qrxn = 831 J + 32.0 J

863 J------Heat of reaction

863 J =(51 g)(Heat Capacity)(25.3 °C - 21.4 °C)

4.34 J/g⋅°C

Okay, so, to solve for this, we're going to have to use q = mcΔT. However, the mass of the calorimeter is not important because not of it is used in the reaction, so really, we are only looking at 2 things, the temperature change and specific heat. So, here is the lightly modified equation we will use:
q = cΔT

Now, just plug in the ΔT we had for the original equation (which was 3.9) and use the specific heat of the calorimeter to get q.

q = (8.20) * (3.9)
q = 31.98, or about 32

The 32 is what was absorbed by the calorimeter, so we can add that to the original value to get our answer.
831 + 32 = 863 J

So, the answer is 863 J

Hope this helped!! :D

Why are cells small?

Answers

Because if a cell grows beyond a certain limit, not enough material will be able to cross the membrane fast enough to accommodate the increased cellular volume.
They are also small so they can maximize the ratio of surface area to volume. Smaller cells have a higher ratio which allows more molecules and ions move across the cell membrane per unit of cytoplasmic volume.

basically cells are small because they need to be able to get the nutrients in and waste out quickly.

Answer:

you know what i really dont know but imagine if they were huge

Explanation:

How many pairs of non-bonding electrons are in a silicon disulfide molecule?

Answers

Silicon disulfide (SiS2) is a covalent compound consisting of silicon (Si) and sulfur (S) atoms. To determine the number of pairs of non-bonding electrons in a silicon disulfide molecule, we can first calculate the total number of valence electrons in the molecule.

Silicon (Si) has 4 valence electrons, and sulfur (S) has 6 valence electrons each. Since there are two sulfur atoms in SiS2, that's a total of 2 x 6 = 12 valence electrons from sulfur.

Now, add the valence electrons from silicon and sulfur:

4 (Si) + 12 (S) = 16 valence electrons in total.

In a covalent compound like silicon disulfide, all the valence electrons are involved in bonding, either in forming bonds (shared electrons) or as non-bonding pairs.

Since SiS2 forms two covalent bonds (Si-S-Si), each involving two electrons, there are 4 electrons used for bonding (2 pairs of bonding electrons). Therefore, the remaining 16 - 4 = 12 electrons are non-bonding pairs.

So, there are 12 pairs of non-bonding electrons in a silicon disulfide (SiS2) molecule.

Lithium has two isotopes.Li-6 and Li-7.
Li-6 mass = 6.015121amu
Li-7 mass = 7.016003amu
Lithium Mass = 6.94049amu

What is percent abundance of Li-6?

I know that there is an equation:
6.9409 = (% Abundance Li-6 * Weight of Li-6) + (% Abundance Li-7 * Weight of Li-7)

I don't know how to use it.

Answers

This is in the assumption that these are the only 2 isotopes. This means that:
abundance of Li-6 = X
abundance of Li-7 = 1-X

Hence from your equations
6.9409 = (X * 6.015121amu) + ((1-X) * 7.016003amu)
Solve for X = 0.075 or 7.50%

Answer: This is in the assumption that these are the only 2 isotopes. This means that:

abundance of Li-6 = X

abundance of Li-7 = 1-X

Hence from your equations

6.9409 = (X * 6.015121amu) + ((1-X) * 7.016003amu)

Solve for X = 0.075 or 7.50%

Explanation:

PLEASE HELPConsider the chemical equations shown here.

3 equations. 1: upper N upper O gas plus upper O subscript 3 gas right arrow upper N upper O subscript 2 gas plus upper O subscript 2 gas. Delta H subscript 1 equals 198.9 kilojoules. 2: StartFraction 3 over 2 EndFraction upper O gas right arrow upper O subscript 3 gas. Delta H subscript 2 equals 142.3 kilojoules. 3: upper O gas right arrow one half upper O subscript 2 gas. Delta H subscript 3 equals negative 247.5 kilojoules.

What is DeltaHrxn for the reaction shown below?

NO(g)+O(g)->NO2(g)

Answers

Answer:

-304.1

I had this question and that’s what I got

Answer:

-867.7

Explanation:

yeah

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