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Virty [35]

3.8mL of 0.42 phosphoric acid is required.

Reaction

2H3PO4 + 3CaCL3 → Ca3(PO)4 + 6HCl

moles CaCl2 =0.16 mol/L x0.010 L = 0.0016 mol

moles of H3Po4

= 0.0016mol of CaCl2 x 2 mole of H3PO4/3mole of CaCl2

= 0.00106 mol

V of H3PO4 = 0.0016/0.42 = 0.0038L = 3.8mL

V of H3PO4=3.8mL

To know more about calculation in milliliters refer to:-

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4 0

2 years ago

Photosynthesis can be represented by6CO2(g)+6H2O(l)⇌C6H12O6(s)+6O2(g)Which of the following will be false when the photosynthesi

denis-greek [22]

Answer:

The concentration of O2 will begin decreasing and The concentrations of CO2 and O2 will be equal.

Explanation:

Equilibrium occurs when the velocity of the formation of the products it's equal to the velocity of the formation of the reactants, thus the concentrations of the compounds remain constant.

Analyzing the information and the reaction given, we can notice that in equilibrium the rate (velocity) of formation of O2 (product) is equal to the rate of formation of CO2 (reactant).

As the CO2 and H2O are placed in the reaction, the Le Chateliêr's principle states that the equilibrium must shift to reestablish the equilibrium, thus, they must be consumed, and the concentration of O2 must increase.

As state above, in equilibrium, the concentrations didn't change, thus, the concentrations of CO2 and O2 will not change.

The concentrations of CO2 and O2 depends on the rate of the reaction and the initial quantities presented, so it's not possible to affirm they'll be equal.

8 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 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

Medical implants and high-quality jewelry items for body piercing are frequently known by a trade name, G23Ti, which means "surg

mrs_skeptik [129]

Answer:

Ti6Al4V1 or Ti6Al4V

Explanation:

So, from the question above we are given the following Important data or parameters or information which is going to help us in solving this question and they are;

The "surgical-grade" titanium is made up of 64.3% titanium, 24.19% aluminum and 11.42% vanadium.

This is a stoichiometry Question so in order to be able to solve this question one must have the basic knowledge of stoichiometry, it is advisable that one revise it.

So, let us delve right into the solution of this question. Recall that, the formula for Calculating the number of moles = mass/molar mass.

Hence, the number of moles of titanium,Ti = 64.3/47.87 = 3 moles.

Also, the number of moles of aluminum = 24.19/26.98 = 0.8966

For vanadium; 11.42/51 = 0.22

The next step is to divide each number of moles component by the lowest number of mole, that is;

Ti = 1.34/ 0.22 = 6; Al = 0.8966/0.22= 4 and V = 0.22/0.22 = 1.

Thus, the emperical formula is = Ti6Al4V.

7 0

2 years ago