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3 years ago

Label each transition in this flow chart as a chemical change or a physical change

Chemistry

1 answer:

Elodia [21]3 years ago

4 0

1) Left up: a chemical change. We can see new substance (red-blue) is formed from one blue and one red atom.

In chemical change new substances are formed, the atoms are rearranged and the reaction is followed by an energy change.

2) Left down: a chemical change. We can see new substance (red-blue) is formed from two blue and one red atoms.

3) MIddle: a physical change. There is no new substance. Bonds are not broken.

4) Right up: a chemical change. Bonds are broken.

5) Right down: a physical change. Change of state of matter.

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The reaction for photosynthesis producing glucose sugar and oxygen gas is:

Anvisha [2.4K]

1.5 grams of glucose is produced from 2.20 g of CO₂.

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:

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

mass of glucose= 1.5 grams

Then, 1.5 grams of glucose is produced from 2.20 g of CO₂.

5 0

3 years ago

PLEASE ANSWER AS QUICKLY AS POSSIBLE THANK YOU SO MUCH

Harlamova29_29 [7]

Answer:

This description needs a unit describing the system of measurement.

6 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$