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2 years ago

What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride

Chemistry

1 answer:

Savatey [412]2 years ago

7 0

Given:

Volume of Na2CO3 = 250 ml = 0.250 L

Molarity of Na2CO3 = 6.0 M

Volume of CaF2 = 750 ml = 0.750 L

Molarity of CaF2 = 1.0 M

To determine:

The mass of CaCO3 produced

Explanation:

Na2CO3 + CaF2 → CaCO3 + 2NaF

Based on the reaction stoichiometry:

1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3

Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles

Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles

CaF2 is the limiting reagent

Thus, # moles of CaCO3 produced = 0.750 moles

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 produced = 0.750 moles * 100 g/mol = 75 g

Ans: Mass of CaCO3 produced = 75 g

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Before landing, the brakes and the tires of an airliner have a temperature of 15.0∘C. Upon landing, the 90.7 kg carbon fiber bra

Goryan [66]

Answer:

0.921 J/g degrees C

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation). Therefore, heat given off by the brakes = −heat taken in by tires, or:

−qbrakes=qtires

The equation used to calculate the quantity of heat energy exchanged in this process is:

−qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires

First we must convert the mass of the tires and the brakes from kg to g.

massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g

masstires=123 kg×1,000. g1 kg=1.23×105 g

Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C.

−(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C)

ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C

The answer should have three significant figures, so round to 0.921Jg∘C.

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2 years ago

When a pH probe is inserted into a solution containing the chloride ion it is neutral. What is the pH of a solution containing t

vivado [14]

Answer:

a. the solution will be weakly basic.

b. Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.

Explanation:

a. The fluoride ion (F⁻) reacts with water thus:

F⁻ + H₂O → HF + OH⁻

That means that fluoride ions produce OH⁻ ions in solution doing the solution will be weakly basic.

b. The acidic equilibrium of NH₄⁺ is:

NH₄⁺ ⇄ NH₃ + H⁺ with a ka of 5,6x10⁻¹⁰.

The basic equilibrium of CN⁻ is:

CN⁻ + H₂O → HCN + OH⁻ with a kb of 2x10⁻⁵

That means that the production of OH⁻ from CN⁻ is higher than production of H⁺ from NH₄⁺. The CN⁻ is a stronger base than NH₄⁺ is an acid.

Thus, the pH of a salt solution of NH₄CN would be Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.

I hope ot helps!

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2 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

Given that 25.0 mL of mercury has a mass of 340.0 g, calculate (a) the density of mercury and (b) the mass of 120.0 mL of mercur

natita [175]

Answer :

(a) The density of mercury is, 13.6 g/ml

(b) The mass of 120.0 ml of mercury is, 1632 grams

Explanation :

(a) Now we have to calculate the density of mercury.

Given :

Volume of mercury = 25.0 ml

Mass of mercury = 340.0 g

Formula used :

$\text{Density of mercury}=\frac{\text{Mass of mercury}}{\text{Volume of mercury}}$

$\text{Density of mercury}=\frac{340.0g}{25.0ml}=13.6g/ml$

Therefore, the density of mercury is, 13.6 g/ml

(b) Now we have to calculate the mass of 120.0 ml of mercury.

As, 25.0 ml of mercury has mass = 340.0 g

So, 120.0 ml of mercury has mass = $\frac{120.0ml}{25.0ml}\times 340.0g=1632g$

Therefore, the mass of 120.0 ml of mercury is, 1632 grams

3 0

2 years ago

Protein x has an absorptivity of 0.4 ml·mg-1 ·cm-1 at 280 nm. What is the absorbance at 280 nm of a 2.0 mg ·ml-1 solution of pro

Evgen [1.6K]

Absorbance measures the ability of the substance to absorb light at a specific wavelength.

Absorbance is also equal to the product of molar absorptivity, path length and molar concentration.

The mathematical expression is given as:

$A= \epsilon l c$ (1)

where, A = absorbance

$\epsilon$ = molar absorptivity

l = path length

c = molar concentration.

The above formula is said to Beer's Law.

Absorptivity of protein x = $0.4 mLmg^{-1}cm^{-1}$

Path length = 1 cm

Molar concentration = $2.0 mg mL^{-1}$

Put the values in formula (1)

$Absorbance at 280 nm = 0.4 mL mg^{-1}cm^{-1}\times 1 cm \times 2.0 mg mL^{-1}$

= $0.8$

Thus, absorbance at 280 nm = $0.8$

3 0

2 years ago