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3 years ago

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an experiment requires that each student use an 8.5 cm length of magnesium ribbon. how many students can do the experiment if th

ere is 570-cm length of magnesium ribbon available

1 answer:

skelet666 [1.2K]3 years ago

4 0

I think about 67 students because if you divide 570 by the amount each person needs you get 67.0588235294 so you have to round down

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For a laboratory investigation some students put a strip of shiny metal into a beaker of blue solution and then stored the beake

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Answer:

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Explanation:

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3 years ago

QUESTION 4 Which structure is found in all eukaryotic cells? large central vacuole Golgi apparatus flagella cilia

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4 years ago

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A 15.0 L tank of gas is contained at a high pressure of 8.20 x 10^4 torr. The tank is opened and the gas expands into an empty c

Inga [223]

Answer:

20.5torr

Explanation:

Given parameters:

V₁ = 15L

P₁ = 8.2 x 10⁴torr

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Unknown:

P₂ = ?

Solution:

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Boyle's law is given mathematically as;

P₁ V₁ = P₂V₂

where P₁ is the initial pressure

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V₂ is final volume

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P₂ = 20.5torr

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3 years ago

Which of the following electron configurations for neutral atoms is correct?

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In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?

dedylja [7]

<u>Answer:</u> The correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

<u>Reaction 1:</u> $2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 2:</u> $CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 3:</u> $HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 4:</u> $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

<u>Reaction 5:</u> $SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

6 0

3 years ago