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mafiozo [28]
4 years ago
6

Compound element ratio of 1:2:6 Ba(NO3)2+Na2SO4

Chemistry
2 answers:
Karo-lina-s [1.5K]4 years ago
5 0

1:2:6? Ba(NO3)2 +Na2SO4 -----> BaSO4 + 2NaNO3

Ba(NO3)2

BaSO4

NaNO 3

Na2SO4

Likurg_2 [28]4 years ago
5 0

Answer:

Ba(NO3)2

Explanation:

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What is the volume of a sample that has a mass of 20 grams and density of 4 g/mL?
balandron [24]

Answer:

<h2>5 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{20}{4}  \\

We have the final answer as

<h3>5 mL</h3>

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7 0
3 years ago
By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling
vova2212 [387]

It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.

By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then

p_{1} =patm+pH_{2} O

p_{1} = 101325+ρgh_{1}

It is given that height is 125ft. Put the value of h in above formula:

h1  =125ft=38.1m

ρ=1.04g/mL=1040kg/m^{3}

g=9.81                                  

p_{1} =101325Pa+388711.44

p_{1}   =490036.44‬Pa                

p_{2} =p atm  =101325Pa

It is known that volume and pressure can be expressed as:

V*P=const.

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/101325

V_{2} /V_{1}=4.84

Assume constant temperature

d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.

now p_{1} =p atm​+pH_{2} O =490036.44Pa

V*p=const                              

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/X

p_{2} =490036.44pa/(V2/V1) =326690.96Pa

p_{2} =patm +pH_{2} O

p_{2} =101325Pa+ρgh_{2}

326690.96Pa=101325Pa+ρgh_{2}

ρgh1  =151987.5-101325=225365.96‬‬Pa

ρ=1,04g/mL=1040kg/m3

g=9.81

h_{2} =225365.96/‬ρ∗g

​h_{2}  =225365.96  / ‬1040∗9.81

h_{2}  =22.09m= 72.47ft

ΔH=H_{1} -H_{2}

=125-72.47

=52.53ft

So she can safely ascend up to 52.53 ft without Breathing out

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Calculate ΔHrxn for the following reaction: 5C(s) 6H2(g)→C5H12(l) Use the following reactions and given ΔH values: C5H12(l) 8O2(
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Answer: its 4xb236.52

Explanation:

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When working with hazardous substances in the laboratory your partner splashes a small amount of Sodium Hydroxide onto their clo
alexira [117]

Answer:

a) After helping our partner, we should immediately report the incident to the lab manager or any person in charge of the emergencies occurring in the lab.

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Aluminum is manufactured using electrolysis. Carbon electrodes are used. Describe the nature of the electrolyte.​
Aleksandr [31]

Answer:

The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.

Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).

However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.

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