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marshall27 [118]
3 years ago
8

Energy + 6CO2 + 6H2O → 6O2 + C6H12O6 The purpose of photosynthesis is to produce usable chemical energy or glucose from solar en

ergy. If the formula for glucose is C6H12O6, how many atoms of carbon are used to produce the glucose?
A) 1

B) 2

C) 6

D) 9
Chemistry
2 answers:
Natalija [7]3 years ago
8 0

Answer:

The answer is c) 6

Explanation:

In the expression of the reaction  

Energy + 6CO2 + 6H2O → 6O2 + C6H12O6

you can see that to produce a mol of glucose (C6H12O6) you need 6 moles of CO2, which it is the reagent that contains the atom of carbon used to produce glucose. Now you multiply the coefficient 6 by the subscript 1, which is the amount of carbon atoms that the molecule CO2 contains. So you end up with <u><em>6 atoms of carbon.</em></u>

slava [35]3 years ago
7 0
C stands for carbon atoms. In glucose thera are 6 carbon atoms, so 6 carbon atoms are used to produce glucose.

Answer: C) 6
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The chemical formula for table sugar is c12h22o11. what can you tell from this formula
guajiro [1.7K]
The chemical formula for table sugar is C12H22O11.  From the formula, you can say that it is an organic molecule where in every molecule contains 12 atoms of carbon, 22 atoms of hydrogen and 11 atoms of oxygen.  There are 45 atoms in total in one molecule of sugar. 
6 0
3 years ago
What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i
Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314  (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0
3 years ago
12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?
Kaylis [27]

Answer:

115.625^{\circ}\text{F}

Explanation:

m_1 = First mass of water = 12 oz

m_2 = Second mass of water = 20 oz

\Delta T_1 = Temperature difference of the solution with respect to the first mass of water = (T-75)^{\circ}\text{F}

\Delta T_2 = Temperature difference of the solution with respect to the second mass of water = (T-75)^{\circ}\text{F}

c = Specific heat of water

As heat gain and loss in the system is equal we have

m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}

The final temperature of the solution is 115.625^{\circ}\text{F}.

3 0
3 years ago
Which valid empercal formula below would have one. Five times as much oxygen as there is gold in the substance
stealth61 [152]

Answer:

Au2O3

Explanation:

See attachment.

3 0
2 years ago
For the reaction
ddd [48]

Answer:

52.2 g

Explanation:

Step 1: Write the balanced equation

3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O

Step 2: Calculate the moles corresponding to 89.7 g of KOH

The molar mass of KOH is 56.11 g/mol.

89.7 g × 1 mol/56.11 g = 1.60 mol

Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH

The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.

Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

0.533 mol × 97.99 g/mol = 52.2 g

3 0
3 years ago
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