All categories

2 years ago

Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)

Chemistry

1 answer:

marin [14]2 years ago

5 0

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

$S.G=\frac{D}{d_w}$

$d_w$ = density of water = 1 g/mL

$D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL$

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = $\frac{Mass}{Density}$

$=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L$

1 mL = 0.001 L

$Molarity = \frac{n}{V(L)}$

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

$Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L$

You might be interested in

Need help asap with this chemistry if someone could help me

Burka [1]

Answer:

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

Structure Number Two would likely be the most stable structure.

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

The N atom is the one that is "likely" to be attracted to an anion. See explanation.

Explanation:

When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:

$\text{Formal Charge} \\ = \text{Number of Valence Electrons in Element} \\ \phantom{=}-\text{Number of Chemical Bonds} \\\phantom{=} - \text{Number of nonbonding Lone Pair Electrons}$ .

For example, for the N atom in structure one of the first question,

N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
This N atom is connected to only 1 chemical bond.
There are three pairs, or 6 electrons that aren't in a chemical bond.

The formal charge of this N atom will be $5 - 1 - 6 = -2$ .

Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.

Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.

Similarly,

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.

Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.

5 0

3 years ago

24 grams of CH4 was added to the above reaction. Calculate the theoretical yield of CO2. A. 66 grams B. 33 grams c. 132 grams. D

Sonbull [250]

Answer:

Option A. 66 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the mass of CH₄ that reacted and the mass of CO₂ produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4 = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Finally, we shall determine the theoretical yield of CO₂. this can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Therefore, 24 g of CH₄ will react to produce = (24 × 44) /16 = 66 g of CO₂.

Thus, the theoretical yield of CO₂ 66 g

8 0

3 years ago

The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38

adell [148]

Answer:

$1.21x10^{-3}$ M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x $10^{-3}$

H = $\frac{0.21}{1.38x10^{-3} }$

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = $1.21x10^{-3}$ M

7 0

3 years ago

In each of the following blanks, only enter a numerical value.

tatiyna

Answer:

1) 1,... 2

2) 18

3) n= 3 and I=1

Explanation:

1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons

2) the maximum number of electron is given by 2n^2= 2×3^2= 18

3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1

5 0

2 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

Best regards!

6 0

2 years ago