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3 years ago

12

Which one has a greatest mass, one atom of carbon, one atom of hydrogen, or one atom of litium

1 answer:

SpyIntel [72]3 years ago

5 0

Lithium, it has an atomic number of about 3

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If you have 3.0 moles of argon gas at STP, how much volume will the argon take up?

Anestetic [448]

If you have 3.0 moles of argon gas at STP u would take up 2.5 volume

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3 years ago

To separate a mixture of iron filings and salt, the most efficient method would be

GaryK [48]

Answer:

Explanation:

Of course you could do the separation chemically. Dissolve the salt up in water, pass thru a filter, wash the iron filings with ethanol, which would encourage the salt to precipitate from solution.

I do hope I helped you! :)

4 0

3 years ago

Are the chemical reactions that take place when a battery connected endothermic or exothermic

Vilka [71]

Answer:

The answer is endothermic as the heat flows into the system from the surroundings. The products are at higher energy than the reactants, as they have absorbed energy.

8 0

2 years ago

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

8 0

3 years ago

A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man

kap26 [50]

<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

Mass of NaBr sample is 11.97 g

% composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

A sample of 11.97 g of NaBr contains 22.34% of Na by mass

Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

= 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0

2 years ago