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Sidana [21]
3 years ago
14

1. Is mass conserved when 50 g of sugar undergoes a physical change? Use complete sentences to support your answer by explaining

how this can be demonstrated.
2. During an investigation a scientist burned 48 g of magnesium strip. After the reaction the total mass of the product formed was found to be 80 g. Does the law of conservation of mass hold true in this case? Use complete sentences to justify your answer based on numerical calculations.

ACTUALLY answer the question please instead of posting stupid comments like on the last one
Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
6 0
The law of conservation of mass applies to both cases, the following will explain how:

1. When 50 g of sugar undergoes a physical change, say for example dissolving or melting, the amount of sugar in the solution or melted form will also be 50 grams. This shows that the mass before and after the change was the same, in accordance to the law of conservation of mass.

2. When the magnesium strip is burned, this is a chemical reaction. The problem is that we only measure the mass of one of the substances involved in the reaction, the magnesium strip, which makes it seem like the mass has increased. Actually, during burning, the magnesium combines with oxygen in the air. This oxygen was present before the reaction, we just did not measure it. And after the reaction it is present in the form of product. Therefore, mass is still conserved.
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A 58.33-g sample of milk of magnesia, Mg(OH)2, always contains 24.31g of magnesium, 32.00g of oxygen, and 2.02g of hydrogen. Fin
ASHA 777 [7]

The mass percentage of each element in the milk of magnesia is, 41.68% Mg, 54.86% Oxygen, and 3.46% Hydrogen elements

Give the total mass of a sample of milk of magnesia Mg(OH)₂ as 58.33 g

We are to calculate the percentage by mass of each element

\% mass =\frac{mass \ of \ element}{mass \ of \ sample} \times 100

For Magnesium:

Mass of magnesium = 24.31g

\% mass \ of \ Mg = \frac{24.31}{58.33} \times 100\\ \% mass \ of \ Mg =\frac{2431}{58.33} \\ \% mass \ of \ Mg = 41.68\%

For oxygen element:

Mass of oxygen = 32.00g

\% mass \ of \ O_2 = \frac{32}{58.33} \times 100\\ \% mass \ of \ O_2 =\frac{3200}{58.33} \\ \% mass \ of \ O_2 = 54.86\%

For the hydrogen element

Mass of hydrogen = 2.02g

\% mass \ of \ H_2 = \frac{2.02}{58.33} \times 100\\ \% mass \ of \ H_2 =\frac{202}{58.33} \\ \% mass \ of \ H_2 = 3.46\%

Hence the mass percentage of each element in the milk of magnesia is, 41.68% Mg, 54.86% Oxygen, and 3.46% Hydrogen elements.

Learn more here: brainly.com/question/20065048

5 0
3 years ago
Using the equation 40kJ+2SO3(g)-->2SO2(g) + O2(g), if pressure is added what way
Ad libitum [116K]

In an equation, 40kJ + 2SO₃ (g) -->2SO₂(g) + O₂(g), if pressure added, the equilibrium shifts to the left side of the reaction because it has fewer moles of gas.

<h3>What is Le Ch atelier's principle?</h3>

Le Ch atelier's principle states that if the pressure is added in a nay reaction, then the equilibrium will be shifted toward the side where there are fewer atoms.

In this case, there are fewer moles on the left side.

Thus, when pressure is added, the equilibrium shift to the left side due to Le Ch atelier's principle.

Learn more about Le Ch atelier's principle

brainly.com/question/2001993

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6 0
2 years ago
What kind of oxide is formed when a piece of sodium is dropped in the water
aleksley [76]

Answer:

Sodium oxide is the product

Explanation:

4Na+O2->2Na2O

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3 years ago
Be sure to answer all parts. Determine the electron-group arrangement, molecular shape, and ideal bond angle for the following m
choli [55]

Answer:

trigonal planar

Explanation:

The molecule SO3 is of the type AX3. The molecule is symmetrical and non polar.

There are three regions of electron density in the molecule. This corresponds to a trigonal planar geometry. This means that the three oxygen atoms are arranged at the corners of a triangle.  The bond angle is 120 degrees.

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Get ready for the wave of screenshots :&gt;
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Answer:

okay let me know when i can help, ill be happy too help

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