How many grams are in 4 moles of H2O?

According to the equation below, how many moles of PbO are required to generate 3.88×1023 nitrogen molecules?

saul85 [17]

Answer:

1.935 mole

Explanation:

We'll begin by calculating the number of mole present in 3.88x10^23 molecules of nitrogen(N2). This can be obtained as follow:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02x10^23 molecules. Therefore 1 mole of N2 contains 6.02x10^23 molecules.

Now if 1 mole of N2 contains 6.02x10^23 molecules,

Then Xmol of N2 will contain 3.88x10^23 molecules i.e

Xmol of N2 = (3.88x10^23)/6.02x10^23

Xmol of N2 = 0.645 mole

Now, we can obtain the number of moles of PbO required to generate 3.88x10^23 molecules (i.e 0.645 mole) of N2. This is illustrated below:

The equation for the reaction is given below:

3PbO + 2NH3 → 3Pb + N2 + 3H2O

From the balanced equation above, 3 moles of PbO produced 1 mole of N2.

Therefore, Xmol of PbO will produce 0.645 mole of N2 i.e

Xmol of PbO = 3 x 0.645

Xmol of PbO = 1.935 mole.

From the calculations made above,

1.935 mole of PbO will produce 3.88x10^23 molecules of nitrogen (N2).

8 0

3 years ago

A sample of nitrogen gas has a volume of 478 cm3 and a pressure of 104.1 kPa. What volume would the gas occupy at 88.2 kPa if th

nadezda [96]

<span>1. </span>To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

V2 = 104.1 x 478 / 88.2

6 0

3 years ago

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Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →

Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = $\frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }$

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0

3 years ago

If a system is left to change spontaneously, in what state will it end?

lesya692 [45]

Answer:

The state with the lowest possible energy consistent with the state of maximum disorder

so letter A

3 0

3 years ago

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Describe how to prepare 10 ml of 5, 10, 15, and 20 micro M CV solution using a 25 microM CV stock solution

zalisa [80]

A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a stock solution whose concentration is known.

4 0

3 years ago

How many grams are in 4 moles of H2O?​

How many grams are in 4 moles of H2O?