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erik [133]
2 years ago
15

Which process produces the energy that is used by solar panels?

Chemistry
2 answers:
Setler [38]2 years ago
8 0

Answer:

A

Explanation: got it right

KengaRu [80]2 years ago
6 0

Answer:

B

Explanation:

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2KMnO4= K2MnO4+ MnO2+O2 how many grams of KMnO4 are required to produce 1.60 grams of O2
Sergeu [11.5K]

Answer: 15.8 g of KMnO_4 will be required to produce 1.60 grams of O_2

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} O_2=\frac{1.60g}{32g/mol}=0.05moles

2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2

According to stoichiometry :

As 1 mole of O_2 is given by = 2 moles of KMnO_4

Thus 0.05 moles of O_2 is given by =\frac{2}{1}\times 0.05=0.10moles  of KMnO_4

Mass of KMnO_4=moles\times {\text {Molar mass}}=0.10moles\times 158g/mol=15.8g

Thus 15.8 g of KMnO_4 will be required to produce 1.60 grams of O_2

5 0
3 years ago
What mass of nitrogen trifluoride can be produced when 32.0 G of fluorine is reaction according to the table below
Sunny_sXe [5.5K]

When 32.0 g of F₂ react with enough N₂, 39.86g of NF₃ are formed.

<h3>What is stoichiometry?</h3>

Stoichiometry is the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical change.

Let's consider the following balanced equation.

N₂ + 3 F₂ ⇒ 2 NF₃

To calculate the mass of NF₃ formed from 32.0 g of F₂, we need to consider the following relationships.

  • The molar mass of F₂ is 38.00 g/mol.
  • The molar ratio of F₂ to NF₃ is 3:2.
  • The molar mass of NF₃ is 71.00 g/mol.

32.00gF_2 \frac{1molF_2}{38.00gF_2} \frac{2molNF_3}{3molF_2} \frac{71.00gNF_3}{1molNF_3} = 39.86gNF_3

When 32.0 g of F₂ react with enough N₂, 39.86g of NF₃ are formed.

Learn more about stoichiometry here: brainly.com/question/16060199

7 0
2 years ago
For the decomposition of ammonia on a platinum surface at 856 °C
Crazy boy [7]

\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}

  • dt remains same for reaction

\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}

\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})

\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}

M is molarity here not metre

5 0
2 years ago
At 250°C an equilibrium mixture of SbCl3(g), Cl2(g), and SbCl5(g) has the partial pressures 0.670 bar, 0.438 bar, and 0.228 bar,
Sveta_85 [38]

Explanation:

As the given reaction will be SbCl_{3}(g) + Cl_{2}(g) \rightarrow SbCl_{5}

Since, it is given that volume of reaction vessel is doubled. Hence, moles of species involved will also increase.

Therefore, the reaction equation will become as follows.

             2SbCl_{3}(g) + 2Cl_{2}(g) \rightarrow 2SbCl_{5}

As it is given that partial pressure of SbCl_{3} is 0.670 bar, Cl_{2} is 0.438 bar and SbCl_{5} is 0.228 bar.

Expression to calculate new equilibrium pressure if the volume of reaction vessel is double is as follows.

                  K_{p} = \frac{[P_{SbCl_{5}}]^{2}}{[P_{SbCl_{3}}]^{2}[P_{Cl_{2}}]^{2}}

                                   = \frac{(0.228)^{2}}{(0.670)^{2}(0.438)^{2}}

                                   = 6.05 \times 10^{3}

Thus, we can conclude that new equilibrium pressures if the volume of the reaction vessel is doubled is 6.05 \times 10^{3}.

4 0
3 years ago
Difference between elimination and substitution​
kotegsom [21]

Answer:

The obvious difference between substitution and elimination reactions is that substitution reactions replace one substituent with another while elimination reactions simply remove the substituent. ... The major difference between the groups is the number of steps the reaction takes place over.

substitution and elimination can be seenis by looking at the bond or the H+ . When substitution occurs, the H+ will be subtituted with other atom. As for elimination, when this reaction occur which mostly involve alkane, new bondwill appear.

Thank you and please rate me as brainliest as it will help me to level up

5 0
3 years ago
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