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3 years ago

Which of these equations best summarizes photosynthesis?

Chemistry

1 answer:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer: C

Explanation:

Sunlight

6 CO2 + 6 H2O --------------> C6H12O6 + 6 O2

Chlorophyll

Carbon(IV)oxide Water Glucose Oxygen

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What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen?

djverab [1.8K]

To get the empirical formula of this compound, we take a basis of 100 grams which means each percentage is equivalent to 1 gram. Hence there is 32.39 grams sodium, 22. 53 grams sulfur and 45.07 grams oxygen. We convert each mass to their moles by dividing by their respective molar mass. Na: 1.408, S:0.704 and O:2.82. divide each with the lowest: Na: 2: S: 1 and O:4. Hence the formula is Na2SO4.

5 0

3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

$m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2$

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

$m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu$

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

$m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3} \\\\m_{AgNO_3}=2.03gAgNO_3$

Best regards!

5 0

3 years ago

What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?

My name is Ann [436]

Answer:

602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The yield is measured by the amount of each product produced by the reaction.

The chemical formula of fluorene is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The oxidation, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

$2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O$

To calculate the yield follow these steps:

1. Mole ratio

$2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O$

2. Convert 175mg of fluorene to number of moles

$175mg/times 1g/1,000mg=0.175g$

Number of moles = mass in grams / molar mass

$\text{number of moles}=0.175g/166.223g/mol=0.0010528mol$

3. Set a proportion for each product of the reaction

a) For CO₂

i) number of moles

$2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x$

$x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2$

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

mass = number of moles × molar mass

mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) For H₂O

i) number of moles

$0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O$

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

mass = number of moles × molar mass

mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg

4 0

3 years ago

How does the line spectra are used to identify elements and what they indicate about atoms?

olga nikolaevna [1]

we know that each element has an unique spectra and it can be used to identify the element. it shows that the energy levels of the electrons and different colors are the result of different wavelengths.
hope it helps

3 0

3 years ago