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4 years ago

Why are hydrogen, lithium, and sodium classified as reactive elements?

1 answer:

4 0

The alkali metals and hydrogen are reactive because they have only one electron to give in order to complete their valence shell. It is easier to give that one electron so when given the opportunity they will. This means they will react with anything polar or willing to take an electron.

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In an experiment Teresa's measures 15.5 mL of water she must have used a

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transfer pipet that had markings every 0.1 mL.

Explanation:

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4 years ago

Which of the following is NOT a testable hypothesis?

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Gravity is caused by a magical force

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4 years ago

Read 2 more answers

How many formula units make up 34.6 g of magnesium chloride (mgcl2)?

PolarNik [594]

The atomic mass of the elements making up MgCl₂ are;
Mg - 24.3
Cl - 35.45
The formula mass of MgCl₂ is 24.3 + (2x35.45) = 95.2
One formula is made of 6.022 x 10²³ formula units of MgCl₂
the formula mass of 95.2 contains 6.022 x 10²³ formula units
Therefore a mass of 34.6 g - $\frac{6.022* 10^{23}*34.6 }{95.2}$
number of MgCl₂ formula units are - 2.19 x 10²³ formula units

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3 years ago

Complete each of the following questions by choosing the correct words

suter [353]

Answer:

to the first

A) mass

to the second

B) Volume

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2 years ago

How many liters of oxygen gas, at standard

Karo-lina-s [1.5K]

Answer:

Explanation:

For the balanced reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

Using cross multiplication:

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

Finally, we can get the volume of oxygen using the information:

It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).

Using cross multiplication:

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

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4 years ago