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photoshop1234 [79]
3 years ago
7

Using the following equation how many grams of water you would get from 886 g of glucose:

Chemistry
1 answer:
ankoles [38]3 years ago
8 0

Answer:

531.6g

Explanation:

Total moles of glucose in this case is: 886/180= 4.922 (mole)

For every 1 mole glucose we get 6 mole water

-> Mole of water is: 4.922 * 6= 29.533 (mole)

weight of water is 18. Therefore, total weight of water that we will have from 886g of glucose are: 25.933*18= 531.6g

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What are specatator ions
Margaret [11]

Answer:

A spectator ion is an ion that does not participate in a chemical reaction and is found in a solution before and after a reaction.

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5 0
2 years ago
Weak acids
maria [59]

Answer:

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3 0
3 years ago
The chemical equation below shows the formation of aluminum oxide (Al2O3) from aluminum (Al) and oxygen (O2).
Rufina [12.5K]

Answer:

Mass = 182.4 g

Explanation:

Given data:

Number of moles of Al₂O₃ = 3.80 mol

Mass of oxygen required = ?

Solution:

Chemical equation:

4Al + 3O₂    →       2Al₂O₃

Now we will compare the moles of aluminum oxide and oxygen.

                Al₂O₃           :           O₂

                   2               :            3

                 3.80            :         3/2×3.80 = 5.7

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 5.7 mol × 32 g/mol

Mass = 182.4 g

4 0
2 years ago
Naoki's bicycle has a mass of 10 kg. If Naoki sits on her bicycle and starts pedaling with a force of 168 N, causing an accelera
White raven [17]

Answer:

50 kg

Explanation:

Data:

Mass of bicycle = 10 kg

                       F = 168 N

                       a = 2.8 m/s²

Calculation:

                     F = ma     Divide each side by m, Then

                 m = F/a

                     = 168/2.8

                      = 60 kg

                 m = mass of bicycle + Naoki's mass. Then

                60 = 10 + Naoki's mass     Subtract 10 from each side

Naoki's mass = 50 kg

3 0
3 years ago
The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0
3 years ago
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