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3 years ago

1.) I am in period two and have an atomic mass of eleven. Who am I?

Chemistry

1 answer:

zavuch27 [327]3 years ago

8 0

1.)Boron

2.)Cadmium has 48 electrons not 121 Mercury has 80 And Copernicum so they all have no 121 electrons

3.)Hydrogen

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When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?

podryga [215]

Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.

c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

3 0

3 years ago

An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.

aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

6 0

3 years ago

25cc of 5 % NaOH solution neutralized 30cc of h2sO4 solution. Whatis normality of H2SO4?

MAVERICK [17]

The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.

<h3>What is the molarity of 5% NaOH?</h3>

The molarity of 5% NaOH is 1.32 M

25 cc of NaOH neutralized 30cc of H₂SO₄ solution.

Equation of reaction is given below:

2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O

Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M

Normality = Molarity × moles of H⁺ ions per mole of acid

moles of H⁺ ions per mole of H₂SO₄ = 2

Normality of H₂SO₄ = 0.55 x 2 = 1.1 N

In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.

Learn more about normality at: brainly.com/question/22817773

#SPJ1

7 0

2 years ago

What system was put into place in 1965 to help people be prepared for future tornados? WILL GIVE BRAINLIEST AND THANKS

Maslowich

Answer:

Eas and NOAA is that system

3 0

3 years ago

"an acid is a substance that forms hydrogen ions when dissolved in water." this is an example of —

LenaWriter [7]

Answer is: this is an example of an Arrhenius acid.

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons (H⁺).

For example hydrochloric acid: HCl(aq) → H⁺(aq) + Cl⁻(aq).

An Arrhenius base is a substance that dissociates in water to form hydroxide ions (OH⁻<span>).
In this example lithium hydroxide is an Arrhenius base:</span>

LiOH(aq) → Li⁺(aq) + OH⁻(aq).

5 0

3 years ago