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Oksanka [162]
3 years ago
7

Find the weight of HNO_3 present in 20ml, 0.30 N

Chemistry
1 answer:
yanalaym [24]3 years ago
6 0

Answer:

mass of HNO₃ = 0.378 g

Explanation:

Normality = Molarity * number of equivalents

Molarity = Normality/number of equivalents

normality of HNO₃ = 0.30 N, Volume = 20 mL

HNO₃ ionizes in the following way:

HNO₃(aq) ----> H⁺ + NO₃⁻

Therefore, number of equivalents for HNO₃ is 1

molarity of HNO₃ = 0.30/1 =0.30 mol/dm³

Using the formula, molarity = number of moles/volume in liters

number of moles = molarity * volume

Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL

number of moles = 0.006 moles

From the formula, mass = number of moles * molar mass

molar mass of HNO₃ = 63.0 g/mol

mass = 0.006 * 63

mass of HNO₃ = 0.378 g

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Boyles Law P1V1 = P2V2
arsen [322]

Answer:

A. The balloons will increase to twice their original volume.

Explanation:

Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:

P ∝ 1/V

P = k/V

PV = k (constant)

P = pressure, V = volume.

P_1V_1=P_2V_2

Let the initial pressure of the balloon be P, i.e. P_1=P, initial volume be V, i.e. V_1=V. The pressure is then halved, i.e. P_2=\frac{P}{2}

P_1V_1=P_2V_2\\\\P*V=\frac{P}{2} *V_2\\\\V_2=\frac{2*P*V}{P}\\\\V_2=2V

Therefore the balloon volume will increase to twice their original volume.

3 0
3 years ago
What is h+ concentrate in a solution with a pH of 1.25 round to nearest hundredth
AlexFokin [52]

Answer:

0.0562

Explanation:

Ph=-log[H+]

to find the h+ is the antilogarithm of the Ph.

Which is 10 raised to the power - Ph.

4 0
2 years ago
What is the mass of one atom of calcium in units of amu?
finlep [7]
40.08 i believe would be it
3 0
3 years ago
13. Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed. (a) Mg(s) + Ni2+(
sp2606 [1]

Answer:

A) Mg ₍s₎ | Mg²⁺₍aq₎║Ni²⁺₍aq₎ | Ni₍s₎

B)  Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

C)  Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

D) Pt₍s₎ | CuNO₃₍ aq₎ , Cu(NO₃)₂₍aq₎ ║ Au(NO₃)₃₍aq₎ | Au₍s₎

Explanation:

A) Mg + Ni²⁺ → Mg²⁺ + Ni

Mg⁰→ Mg²⁺ + 2e⁻    oxidation   anode

Ni²⁺ + 2e⁻ → Ni⁰       reduction   cathode

Mg ₍s₎ | Mg²⁺₍aq₎║Ni²⁺₍aq₎ | Ni₍s₎

anode        salt bridge       cathode

B)   2 Ag⁺₍aq₎ + Cu₍s₎ → Cu²⁺₍aq₎ + 2 Ag₍s₎

2 Ag⁺ + 2e⁻ → 2 Ag⁰   cathode

Cu →  Cu²⁺ + 2e⁻          anode

 Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

anode       salt bridge       cathode

C) Mn ₍s₎  + Sn(NO₃)₂ ₍aq₎ → Mn (NO₃)₂ ₍aq₎  + Au₍s₎

Mn⁰ → Mn²⁺+ 2e⁻       anode

Sn²⁺ + 2e⁻ → Sn⁰₍s₎      cathode

Mn₍s₎ | Mn(NO₃)₂ ₍aq₎ ║ Sn(NO₃)₂ ₍aq₎ | Sn₍s₎

anode                salt bridge                   cathode

D) 3CuNO₃₍ aq₎ + Au(NO₃)₃₍aq₎  → 3Cu(NO₃)₂₍aq₎  + Au₍s₎

3Cu⁺ → 3Cu²⁺ + 3e⁻  anode

Au³⁺ + 3e⁻ → Au⁰                cathode

Pt₍s₎ | CuNO₃₍ aq₎ , Cu(NO₃)₂₍aq₎ ║ Au(NO₃)₃₍aq₎ | Au₍s₎

       anode                           salt bridge                   cathode

Semi-reactions are noted to determine oxidation and reduction. The oxidation reaction occurs at the anode and the reduction reaction at the cathode. In the case of subsections a, b, c the electrodes may be the solid metal, while in the case d the species are in an accusative medium and require the platinum electrode.

5 0
3 years ago
WILL MAKE BRAINLIEST IF ANSWERED ASAP
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The 3rd one is right
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