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3 years ago

H202-h20-o2 In this process oxygen gas is a

Chemistry

1 answer:

Advocard [28]3 years ago

6 0

Answer:

Oxgeyn gas liquid.

Explanation:

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A dose of aspirin of 5.0 mg per kilogram of body weight has been prescribed to reduce the fever of an infant weighing 8.5 pounds

Tanya [424]

<h3><u>Answer</u>;</h3>

19 mg

<h3><u>Explanation</u>;</h3>

dose = 5.0 mg/kg

weight of infant = 8.5 pounds

Convert 8.5 pounds to kg: 8.5 pounds = 3.86 kg

Therefore;

The dose to give the infant is:

= (5.0 mg/kg) x (3.86 kg)

= 19.3 mg

<u>≈ 19 mg</u>

5 0

3 years ago

Read 2 more answers

ILL GIVE YOU BRAINLIEST!!The type of biome that we live in here in Knightdale, NC is what? *

AleksandrR [38]

Answer:

Temperate Deciduous Forest

Explanation:

3 0

3 years ago

A mole is known by what other name?

Margarita [4]

Answer: avogadros number/constant

Explanation:

3 0

2 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

$E^0_(Au^{3+}/Au)$ = 1.40 V

$E^0{cell}=E^0{cathode}-E^0{anode}=1.40-(-0.74)=2.14V$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}$

where,

n = number of electrons in oxidation-reduction reaction = 3

$E^o_{cell}$ = standard electrode potential = 2.14 V

$E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}$

$E_{cell}=2.14$

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.

6 0

4 years ago

When thermal

inn [45]

I just had it it’s was c

8 0

3 years ago