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3 years ago

(Dear brainly, you can't delete this question caue its about school work too)

Chemistry

2 answers:

NeTakaya3 years ago

5 0

Answer:

70% to 100% is the highest grade, a mark of Distinction. 60% to 69% earns a Merit. 50% to 59% is Pass.is what im guessing and if your grade is 90.5 and has lowerd or hasn't that is ok because it might lower to a c but i advise to work better and if it is just a 90.5 overall including the test score that is fine nbut you will pass is what im think

Explanation:

Try better next time :) hope i help so you pass im ptretty sure and if you dont just be a tryhard

STatiana [176]3 years ago

4 0

I don’t understand your question but It looks like the another person answered it already did

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drek231 [11]

Answer:

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3 years ago

What's the electron configuration of an N⁻³ ion

tatuchka [14]

The electronic configuration of n - 3 ion is [He] 2s2 2p3.

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2 years ago

A "fizz-keeper" is a small air pump that you can place on an opened soda bottle. When operated the pump is used to pressurize th

AnnZ [28]

Answer: increase

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

As soda contains carbon dioxide dissolved in water.

The equation given by Henry's law is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$C_{CO_2}$ = solubility of carbon dioxide in water

$K_H$ = Henry's constant

$p_{CO_2}$ = partial pressure of carbon dioxide

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3 0

3 years ago

Calculate the relative density of carbon monoxide and carbon dioxide with respect to air.

JulsSmile [24]

Density is proportional to molar mass, assuming pressure and temperature remain constant. Therefore, since CO has a molar mass of 28 and CO2 has a molar mass of 44:
The relative density of CO vs air is 28/29 = 0.9655.
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3 0

3 years ago

A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+.

Vikki [24]

Answer: a) Anode: $Cr\rightarrow Cr^{3+}+3e^-$

Cathode: $Au{3+}+3e^-\rightarrow Au$

b) Anode : Cr

Cathode : Au

c) $Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E_{cell}=2.14V$

Explanation: -

a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

At cathode: $Au{3+}+3e^-\rightarrow Au$

At anode: $Cr\rightarrow Cr^{3+}+3e^-$

b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.

At anode which is a negative terminal, oxidation occurs which is loss of electrons.

Gold acts as cathode ad Chromium acts as anode.

c) Overall balanced equation:

At cathode: $Au{3+}+3e^-\rightarrow Au$ (1)

At anode: $Cr\rightarrow Cr^{3+}+3e^-$ (2)

Adding (1) and (2)

$Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E^0_(Cr^{3+}/Cr)$ = -0.74 V