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3 years ago

This is science what about this one? According to modern atomic theory, _____ can't move from one energy

1 answer:

5 0

According to modern atomic theory, electrons can't move from one energy
level to another. It is not that it cannot really move but electrons tend to stay at a level with the lowest energy possible. Hope this answers the question. Have a nice day.

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A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

Answer: The mass of glucose in final solution is 1.085 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

Pls help lollollolljdjdjdjdjjdjjzjzjsjsjsis

HACTEHA [7]

Answer:

djhdhdjdududddhhrhed

3 0

2 years ago

A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

5 0

3 years ago

Determine the correct set of coefficients to balance the chemical equation. __c6h6(l) + __o2(g)? __co2(g) + __h2o(g)

forsale [732]

To balance this equation, first we should consider balancing C because it only presents in one reactant and one product. Assuming the coefficient of C6H6 is 1, there are 6 C's in the reactant, so it generates 6CO2. Then consider balancing H for the same reason. If the coefficient of C6H6 is 1, there are 6 H's in the reactant, so it generates 3H2O.
Now that the coefficient of the products are determined, we can balance O. There are 6*2=12 O's in CO2 and 3*1=3 O's in H2O. So the total number of O in the products is 12+3 = 15. O2 is the only reactant that contains O, so to balance the equation, the coefficient of O2 should be 15/2.
Now the equation looks like:
C6H6 + 15/2O2 ⇒ 6CO2 + 3H2O.
Times both sides of the equation by 2 results the final answer:
2C6H6 + 15O2 ⇒ 12CO2 + 6H2O

3 0

3 years ago

Why is ginger used in food preservation

Vaselesa [24]

Spices like ginger are significantly used in preservatives because they have antimicrobial properties in their chemical compounds. They also have antioxidant properties that allow food to be preserved.

6 0

3 years ago