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Marina86 [1]
4 years ago
10

8.00ml of 1.25M lithiukm hydroxide is reacted with sulfuric acid. It is found that 52.87mL of the sulfuric acid is required to c

ompletely neutralize the lithium hydroxide. What is the approximate molarity of sulfuric acid?
Chemistry
1 answer:
BabaBlast [244]4 years ago
8 0

<u>Answer:</u> The molarity of sulfuric acid is 0.0946 M

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is LiOH

We are given:

n_1=2\\M_1=?M\\V_1=52.87mL\\n_2=1\\M_2=1.25M\\V_2=8.00mL

Putting values in above equation, we get:

2\times M_1\times 52.87=1\times 1.25\times 8.00\\\\M_1=\frac{1\times 1.25\times 8.00}{2\times 52.87}=0.0946M

Hence, the molarity of sulfuric acid is 0.0946 M

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Answer:

c

Explanation:

I hope this may help you guys

6 0
2 years ago
A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r
Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

P\cdot V = n\cdot R\cdot T,

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of gas particles in the gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

V = \dfrac{n \cdot R \cdot T}{P}.

Both the temperature of the gas, T, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

  • T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K},
  • T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}.

The volume of the gas is proportional to its temperature if both n and P stay constant.

  • n won't change unless the balloon leaks, and
  • consider P to be constant, for calculations that include T.

V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

Now, keep the temperature at T_1 =258.65\;\text{K} and change the pressure on the gas:

  • P_1 = 0.995\;\text{atm},
  • P_2 = 0.720\;\text{atm}.

The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}

(3 sig. fig. as in the question.).

See if you get the same result if you hold T constant, change P, and then move on to change T.

6 0
3 years ago
Three isotopes of silicon have mass numbers of 28,29, and 30 with an average atomic mass of 28.086 what does this say about the
Sedbober [7]

Answer is: silicon isotope with mass number 28 has highest relative abundance, this isotope is the most common of these three isotopes.

Ar₁(Si) = 28; the average atomic mass of isotope ²⁸Si.  

Ar₂(Si) =29; the average atomic mass of isotope ²⁹Si.  

Ar₃(Si) =30; the average atomic mass of isotope ³⁰Si.  

Silicon (Si) is composed of three stable isotopes, ₂₈Si (92.23%), ₂₉Si (4.67%) and ₃₀Si (3.10%).

ω₁(Si) = 92.23%; mass percentage of isotope ²⁸Si.  

ω₂(Si) = 4.67%; mass percentage of isotope ²⁹Si.

ω₃(Si) = 3.10%; mass percentage of isotope ³⁰Si.

Ar(Si) = 28.086 amu; average atomic mass of silicon.  

Ar(Si) = Ar₁(Si) · ω₁(B) + Ar₂(Si) · ω₂(Si)  + Ar₃(Si) · ω₃(Si).  

28,086 = 28 · 0.9223 + 29 · 0.0467 + 30 · 0.031.

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Define isoelectronic
Pani-rosa [81]

Answer: Isoelectronic means having the same numbers of electrons or the same electronic structure.

Explanation:

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The physical and chemical properties of carbon depend on the crystalline structure of the element. Its density fluctuates from 2.25 g/cm³ (1.30 ounces/in³) for graphite and 3.51 g/cm³ (2.03 ounces/in³) for diamond. The melting point of graphite is 3500ºC (6332ºF) and the extrapolated boiling point is 4830ºC (8726ºF).

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