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tamaranim1 [39]
2 years ago
12

How many grams of water can be produce from the complete reaction of excess nitric acid and 33.2 mL of 0.245 M lithium hydroxide

solution, assuming that lithium hydroxide is the limiting reactant?
Chemistry
1 answer:
Mariulka [41]2 years ago
3 0

Mass of water produced : 0.146 g

<h3>Further explanation</h3>

Given

33.2 mL of 0.245 M lithium hydroxide

Required

mass of water

Solution

Reaction

HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)

mol LiOH :

= M x V

= 0.245 x 33.2 ml

= 8.134 mmol

From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol

mass H₂O :

= mol x MW

= 8.134 x 10⁻³ mol x 18 g/mol

= 0.146 g

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sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the press
Anna [14]

Answer:

-81.5 degrees C or 191.5 K

Explanation:

We want to use Charles' gas law: V/T = V/T

Our initial volume is 3.20 L, and our initial temperature is 125 degrees C, or 125 + 273 = 398 degrees Kelvin.

Our new Volume is 1.54 L, but we don't know what the temperature is. So, we use the equation:

3.20 L / 398 K = 1.54 L / T ⇒ Solving for T, we get: T = 191.5 K

If we want this in degrees Celsius, we subtract 273: 191.5 - 273 = -81.5 degrees C

3 0
3 years ago
Read 2 more answers
A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th
gayaneshka [121]

Answer:

moles Ne = 0.154 mol

moles F₂ = 0.217 mol

Explanation:

Step 1: Data given

Volume of the vessel system = 2.5 L

Total pressure = 3.32 atm at 0.0 °C

The mixture is heated to 15.0 °C

The entropy of the mixture increases by 0.345 J/K

The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R

Step 2: Define the gas

Neon is a monoatomic gas, composed of Ne atoms

 ⇒ Cv(Ne) ≅ (3/2)R

Fluorine is a diatomic gas, composed of F₂ molecules.  

⇒ Cv(F₂) ≅ (5/2)R

Step 3: Calculate moles of gas

p*V = n*R*T

⇒ with p = the total pressure = 3.32 atm

⇒ with V = the total volume = 2.5 L

⇒ with n = the number of moles of gas

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,  

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,  

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

 ⇒ Cv = 17.43 J/mol*K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K

 ⇒ 20.875 – 8.314 * X = 17.43

X = 0.415 , 1 – X = 0.585

moles Ne = (0.415)(0.3703 mol) = 0.154 mol

moles F₂ = (0.585)(0.3703 mol) = 0.217 mol

4 0
3 years ago
Jim went for a 3.75 hour run. How many minutes did he run?
tensa zangetsu [6.8K]
An hour is 60 minutes. So you multiple the hot by 60. You get 180 minutes plus the 75 remaining minutes. The answer is 255
5 0
3 years ago
Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube
Bess [88]

Let's consider the neutralization reaction between HCl and NaOH.

NaOH + HCl ⇒ NaCl + H₂O

To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.

NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol

HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol

Now, let's determine the reactant in excess and the remaining moles of that reactant.

                    NaOH    +     HCl ⇒ NaCl + H₂O

Initial           0.0200       0.0188

Reaction    -0.0188       -0.0188

Final         1.20 × 10⁻³          0

The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³  moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:

[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M

NaOH is a strong base according to the following equation.

NaOH ⇒ Na⁺ + OH⁻

The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02

We will calculate the pH using the following expression.

pH = 14.00 - pOH = 14.00 - 2.02 = 11.98

The pH is 11.98. Since pH > 7, the solution is basic.

You can learn more about neutralization here: brainly.com/question/16255996

7 0
2 years ago
The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance b
SSSSS [86.1K]

Answer:

the stronger light 5.5 m apart from the total illumination​

Explanation:

From the problem's statement , the following equation can be deducted:

I= k/r²

where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality

denoting 1 as the stronger light and 2 as the weaker light

I₁= k/r₁²

I₂= k/r₂²

dividing both equations

I₂/I₁ = r₁²/r₂²=(r₁/r₂)²

solving for r₁

r₁ = r₂ * √(I₂/I₁)

since we are on the line between the two light​ sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus

r₂ = r₁ + d

then

r₁ = (r₁ + d)* √(I₂/I₁)

r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)

r₁*(1-√(I₂/I₁)) =  d*√(I₂/I₁)

r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁))  =

r₁ = d/[√(I₁/I₂)-1)]

since the stronger light is 9 times more intense than the weaker

I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3

then since d=11 m

r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m

r₁ = 5.5 m

therefore the stronger light 5.5 m apart from the total illumination​

5 0
3 years ago
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