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ICE Princess25 [194]
2 years ago
3

Assume that you want to construct a voltaic cell that uses the following half reactions:

Chemistry
1 answer:
disa [49]2 years ago
7 0
I think it’s A cause they should add enough
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How many moles of O are in 2.45 moles of H2CO3?​
Irina-Kira [14]

Answer:

7.35 moles of oxygen

Explanation:

First of all, for 1 mole of H₂CO₃ we have 3 moles of oxygen (can be deduced from the chemical formula of the acid), then the moles of oxygen in 2.45 mole of the compound, which are given in the question, from the carbonic acid will be:  

If in        1 mole of H₂CO₃ we have 3 moles of oxygen

The in    2.45 moles of H₂CO₃ we have X moles of oxygen

X = ( 3 × 2.45 ) / 1 = 7.35 moles of oxygen

3 0
2 years ago
Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.
Mekhanik [1.2K]

Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

                Since you are not given a chemical reaction, just base your solution to the chemical formula given.

Molar mass of Al2(SO4)3 = 342.15 g/mol

7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol O2)(32g O2/1mol O2) = 4.04 g O2

5 0
2 years ago
Read 2 more answers
Watch the video to determine which of the following relationships correctly depict the relationship between pressure and volume
AnnZ [28]

Answer : The correct options are,

(B) V\propto \frac{1}{P}

(C) P\propto \frac{1}{V}

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

V\propto \frac{1}{P}

The relation between the pressure and volume of two gases are:

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas

P_2 = final pressure of gas

V_1 = initial volume of gas

V_2 = final volume of gas

5 0
2 years ago
How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K
Olenka [21]

I’m not sure I need help with this question

6 0
2 years ago
Student Z completes his/her calibration step for part 3 and finds an average massA to be 95.237 g ± 0.005 g. However, in his/her
Troyanec [42]
It is going to be too low because the mass mistakenly used is lower than the initial.
7 0
2 years ago
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