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4 years ago

5

Hurry I will mark brainest if correct!!!! hurry

1 answer:

Sholpan [36]4 years ago

4 0

Answer:

i think the answer might be A the frist one

Explanation: because the chemical started immediately when the girl put in the two red colored and it started turning blue and by eight minutes it turned blue completely.

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you need to prepare 250.0 mL of a 1.50 M HCl solution from a 2.00 M HCl stock solution. assuming that the accuracy of the concen

Savatey [412]

Answer:

I should use a volumetric flask.

Explanation:

If the accuracy of the concentration is important, we need to use a volumetric flask.

7 0

3 years ago

Read 2 more answers

In addition to reacting with gold (see Problem 21.21), aqua regia is used to bring other precious metals into solution. Balance

suter [353]

Aqua regia is used to bring precious metals into the solution .

$3Pt(s)+ 4NO^{-} _{3} (aq)+18 Cl^{-} (aq) +16H^{+} =3Pt Cl^{2-} _{6} (aq) +4NO(g) +8 H_{2} O$

This is the balanced reaction of Pt (platinum) .

<h3>What is aqua regia ?</h3>

It is a mixture of concentrated nitric acid and hydrochloric acid .there is three part of concentrated HCl and 1 part of concentrated nitric acid .

it is red or yellowish liquid . it is extremely corrosive and can cause skin burns . it is frequently used to dissolve gold and platinum .

The equation of reaction of aqua regia with gold :

$Au + 4H^{+} +NO^{-} _{3} +4Cl^{-} =AuCl^{-} _{4} +NO +2H_{2} O$

Learn more about aqua regia here :

brainly.com/question/12982022

#SPJ4

6 0

2 years ago

A 1.362 g sample of an iron ore that contained Fe3O4 was dissolved in acid with all of the iron being reduced to iron (II). The

Natasha2012 [34]

Answer:

a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

b. 18.17% of Fe in the sample

Explanation:

a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:

Fe²⁺ → Fe³⁺ + 1e⁻

MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O

5 times the iron and suming the manganese reaction:

MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O

<h3>MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O</h3>

b. Moles of permanganate in the titration are:

0.03942L × (0.0281 moles / L) = <em>1.108x10⁻³ moles of MnO₄⁻</em>

Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:

1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =

<em>4.431x10⁻³ moles of Fe²⁺</em>. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:

4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =

<h3>0.2474g of Fe you have in your sample.</h3>

Percent mass is:

0.2474g Fe / 1.362g sample ₓ 100 =

<h3>18.17% of Fe in the sample</h3>

6 0

4 years ago

the internal pressure of a cylinder with one ounce of liquid r410a which has been sitting in 100 degree truck

Novay_Z [31]

Yes it is true. Of course it is right.

5 0

3 years ago

The half-life of Erbium-165 is 10.4 hours. After 24 hours a sample has been reduced to a mass of 2 mg. What was the initial mass

Pachacha [2.7K]

Answer:

$[A_0]=10\ mg$

$[A_t]=0.08245\ mg$

Explanation:

Given that:

Half life = 10.4 hours

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{10.4}\ hour^{-1}$

The rate constant, k = 0.06664 hour⁻¹

Time = 24 hours

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t = 2 mg

$[A_0]$ is the initial concentration = ?

So,

$2\ mg=[A_0]\times e^{-0.06664\times 24}$

$[A_0]=10\ mg$

Now, time = 3 days = 3*24 hours = 72 hours ( 1 day = 24 hours)

$[A_0]=10\ mg$

Thus,

$[A_t]=10\times e^{-0.06664\times 72}\ mg=0.08245\ mg$

6 0

4 years ago