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Tanzania [10]
3 years ago
5

giving the incomplete equation 2N2O5 (g) which set of products completes and balances the incomplete equarion

Chemistry
1 answer:
a_sh-v [17]3 years ago
7 0

Answer:

The complete question is as follows

Given the incomplete equation: 2 N2O5(g) ==>  Which set of products completes and balances the incomplete equation?

A)2 N2(g) + 3 H2(g)

B)2 N2(g) + 2 O2(g)

C)4 NO2(g) + O2(g)

D)4 NO(g) + SO2(g)

The correct option is C) 4NO2(g) + O2(g)

Explanation:

Note that the products should be NO2 and O2 since the reactant is entirely made up of N and O. option A is not correct as hydrogen cannot emerge as a product in this reaction. Matter can never be created or be destroyed bu can only change in a chemical reaction. Option D is not also correct for the same reason.

Option B is not correct since it did not balance the number of atoms of O and N in the reactant side of the equation.

The option C) 4NO2(g) + O2(g) is therefore the right option since it balances both the elements and the number of atoms of the elements present.

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A scientist performs an experiment in which a radium–226 (Ra–226) atom decays into radon–222 (Rn–222). The scientist concludes t
tangare [24]

Answer:

B)

Explanation:

5 0
3 years ago
Read 2 more answers
100.0 mL of Ca(OH)2 solution is titrated with 5.00 x 10–2 M HBr. It requires 36.5 mL of the acid solution for neutralization. Wh
miskamm [114]

Answer:

The number of moles HBr = 0.001825

The concentration of Ca(OH)2 = 0.009125 M

Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

Molarity of HBr = 5.00 * 10^-2 M

Volume of HBR = 36.5 mL = 0.0365 L

Step 2: The balanced equation

Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Step 3: Calculate molarity of Ca(OH) 2

b*Va* Ca = a * Vb*Cb

⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

⇒with a = the coefficient of Ca(OH)2 = 1

⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

3 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
Which of the following pairs consists of a weak acid and a strong base? (1 point)
notka56 [123]

Answer:

acetic acid, sodium hydroxide

Explanation:

A strong acid is an acid that ionizes in water to give all its hydrogen ion. Weak acid only ionize to a certain degree. Acetic acid (CH3COOH) only ionize to give one hydrogen ion despite having other hydrogen atom. This account for its weak nature as an acid as shown below:

CH3COOH <=> H^+ + CH3COO^-

A strong base is a base that ionizes in water to give all it hydroxide ion. Sodium hydroxide(NaOH) ionizes to give all its hydroxide ions. This make it a strong base as shown below;

NaOH <=> Na^+ + OH^-

8 0
3 years ago
A monatomic ideal gas that is initially at 1.50 * 105 Pa and has a volume of 0.0800 m3 is compressed adiabatically to a volume o
lubasha [3.4K]

Answer:

300000Pa or 3×10^5 Pa

Explanation:

Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.

Using Boyle's law

P1V1 = P2V2

P1 is the initial pressure = 1.5×10^5Pa

V1 is the initial volume = 0.08m3

P2 is the final pressure (required)

V2 is the final volume = 0.04 m3

From the formula, P2 = P1V1/V2

P2 = 1.5×10^5 × 0.08 ÷ 0.04

= 300000Pa or 3×10^5 Pa.

8 0
3 years ago
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