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Alex Ar [27]
3 years ago
7

PLEASE HELP URGENT WILL GOVE BRANLIEST WITH EXPLANATION

Chemistry
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

no because it already had it in there and it exploded by a fuse

Explanation:

Hope this helps?? :))

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The reaction for photosynthesis producing glucose sugar and oxygen gas is:
Anvisha [2.4K]

<u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

  • CO₂: 6 moles
  • H₂O: 6 moles
  • C₆H₁₂O₆: 1 mole
  • O₂: 6 moles

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}

<em>mass of glucose= 1.5 grams</em>

Then, <u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>

5 0
3 years ago
Which term indicates how strongly an atom attracts the electrons in a chemical bond
Dmitrij [34]
Electronegativity is your answer.

6 0
3 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
2. Why can we eliminate the claim you circled?
Luda [366]

Answer: theres no image or claim

7 0
3 years ago
Read 2 more answers
The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)
valkas [14]

The production of CO_{2} is 3.3480\times 10^{4} tons/day. Converting mass into kg,

1 ton=907.185 kg, thus,

3.3480\times 10^{4} tons=3.037\times 10^{7} kg

Thus, production of CO_{2} will be 3.037\times 10^{7} kg/ day.

The specific volume of CO_{2} is 0.0120 m^{3}/kg.

Volume of CO_{2} produced per day can be calculated as:

V=Specific volume\times mass

Putting the values,

V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day

Thus, volume of CO_{2} produced per year will be:

V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year

Thus, in 4 year volume of CO_{2} produced will be:

V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}

8 0
3 years ago
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