What did ancient people think of lunar eclipses

10. A 2.36-gram sample of NaHCO3 was completely decomposed in an

Airida [17]

Answer:

0.79 g

Explanation:

Let's introduce a strategy needed to solve any similar problem like this:

Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.

Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:

$m_r=m_{NaHCO_3}=2.36 g$

We have two products formed, sodium carbonate and carbonic acid. This implies that the total mass of the products is:

$m_p=m_{Na_2CO_3}+m_{H_2CO_3}$

Apply the law of mass conservation:

$m_r=m_p$

Substitute the given variables:

$m_{NaHCO_3}=m_{Na_2CO_3}+m_{H_2CO_3}$

Rearrange for the mass of carbonic acid:

$m_{H_2CO_3}=m_{NaHCO_3}-m_{Na_2CO_3}=2.36 g - 1.57 g=0.79 g$

8 0

3 years ago

"calculate the ratio of the velocity of hydrogen molecules to the velocity of carbon dioxide molecules at the same temperature"

Ne4ueva [31]

Answer: 1:4.69

Explanation:

The ratio can be expressed as:

Ua/Ub= √(Mb/Ma)

Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2

Mb is the molecular mass of CO2 = 44

Therefore

Ua/Ub= √(44/2)

Ua/Ub = 4.69

Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69

which implies hydogen is about 4.69 times faster than carbon dioxide.

8 0

3 years ago

1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →

Leno4ka [110]

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

<h3>What is Limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

As 80 g NaOH produces 106 g of Na₂CO₃.

Therefore 5 g NaoH will produce ;

106 / 80 x 5 = 6.625 g

Learn more about limiting reagent here ;

brainly.com/question/11848702

#SPJ1

5 0

1 year ago

Find the mass of 175.4mL of benzene if the density is 0.8786g/ml

bulgar [2K]

Hey there!:

Volume = 175.4 mL

density = 0.8786 g/mL

mass = ?

Therefore:

D = m / V

0.8786 = m / 175.4

m = 0.8786 * 175.4

m = 154.10644 g

hope this helps!

5 0

3 years ago

KFell Fe"(CN), + e + Nat → KNaFe'Fe(CN)6

Alinara [238K]

Answer:

Most common oxidation states: +2, +3

M.P. 1535º

B.P. 2750º

Density 7.87 g/cm3

Characteristics: Iron is a gray, moderately active metal.

Characteristic reactions of Fe²⁺ and Fe³⁺

The [Fe(H2O)6]3+ ion is colorless (or pale pink), but many solutions containing this ion are yellow or amber-colored because of hydrolysis. Iron in both oxidation states forms many complex ions.

Aqueous Ammonia

Aqueous ammonia reacts with Fe(II) ions to produce white gelatinous Fe(OH)2, which oxidizes to form red-brown Fe(OH)3:

Fe2+(aq)+2NH3(aq)+3H2O(l)↽−−⇀Fe(OH)2(s)+2NH+4(aq)(1)

Fe3appt.gif

Aqueous ammonia reacts with Fe(III) ions to produce red-brown Fe(OH)3:

Fe3+(aq)+3NH3(aq)+3H2O(l)↽−−⇀Fe(OH)3(s)+3NH+4(aq)(2)

Fe3bppt.gif

Both precipitates are insoluble in excess aqueous ammonia. Iron(II) hydroxide quickly oxidizes to Fe(OH)3 in the presence of air or other oxidizing agents.

Sodium Hydroxide

Sodium hydroxide also produces Fe(OH)2 and Fe(OH)3 from the corresponding oxidation states of iron in aqueous solution.

Fe2+(aq)+2OH−(aq)↽−−⇀Fe(OH)2(s)(3)

Fe4appt.gif

Fe3+(aq)+3OH−(aq)↽−−⇀Fe(OH)3(s)(4)

Fe4bppt.gif

Neither hydroxide precipitate dissolves in excess sodium hydroxide.

Potassium Ferrocyanide

Potassium ferrocyanide will react with Fe3+ solution to produce a dark blue precipitate called Prussian blue:

K+(aq)+Fe3+(aq)+[Fe(CN)6]4−(aq)↽−−⇀KFe[Fe(CN)6](s)(5)

Fe5a1ppt.gif

With Fe2+ solution, a white precipitate will be formed that will be converted to blue due to the oxidation by oxygen in air:

2Fe2+(aq)+[Fe(CN)6]4−(aq)↽−−⇀Fe2[Fe(CN)6](s)(6)

Fe5a2ppt.gif

Many metal ions form ferrocyanide precipitates, so potassium ferrocyanide is not a good reagent for separating metal ions. It is used more commonly as a confirmatory test.

Potassium Ferricyanide

Potassium ferricyanide will give a brown coloration but no precipitate with Fe3+. With Fe2+, a dark blue precipitate is formed. Although this precipitate is known as Turnbull's blue, it is identical with Prussian blue (from Equation 5).

K+(aq)+Fe+2(aq)+[Fe(CN)6]3−(aq)↽−−⇀KFe[Fe(CN)6](s)(7)

Fe5b.gif

Potassium Thiocyanate

KSCN will give a deep red coloration to solutions containing Fe3+:

Fe+3(aq)+NCS−(aq)↽−−⇀[FeNCS]+2(aq)(8)

Fe5cppt.gif

No Reaction

Cl−, SO2−4

7 0

2 years ago