What will happen to a iron nail kept in water for some day

How many moles are in 91.5 grams of Helium?

konstantin123 [22]

In order to find the number of moles with a given mass of Helium, we need to use its molar mass, which is 4.0026g/mol, therefore we will have:

4.0026g = 1 mol of Helium

91.5g = x moles of Helium

x = 22.86 moles of Helium in 91.5 grams

4 0

1 year ago

Please HELP, ASAP! Homework due tomorrow and I need some help.

sweet-ann [11.9K]

Ions are atoms with a charge other than zero. In a neutral atom, the number of protons (positively charged particles) in the nucleus equals the number of electrons orbiting the nucleus.
Atoms can gain or lose electrons (not protons) resulting in a net charge other than zero. Atoms which lose electrons (usually metals) become positively charges, and atoms which gain electrons (usually nonmetals) become negatively charged.

3 0

3 years ago

Write the balanced molecular and net ionic equations for the reaction between aluminum metal and silver nitrate. identify the ox

asambeis [7]

Well in this case, silver nitrate is reduced:

Ag+ + e− → Ag(s) ↓

Meanwhile, the aluminum is oxidized forming a positive ion:

Al(s) → Al3+ + 3e−

To get the overall reaction, we add the half equations so that the electrons are eliminated:

Al(s) + 3Ag+ → Al3+ + 3Ag(s)

And similarly:

Al(s) + 3AgNO3(aq) → Al(NO3)3(aq) + 3Ag(s)

4 0

3 years ago

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An unsaturated solution is formed when 80 grams of a salt is dissolved in 100 grams of water at 40. This salt could be ?

LekaFEV [45]

A simple way to go about this is that we look at the solubility curve, on the x axis we first look at the temperature and then the corresponding value of solute/100g H2O on the y axis, from the 4 curves above only NaNO3 has a curve that can accommodate 80g of salt at 40 without being Saturated since at 40 degrees it can accommodate 105g of salt to become completely Saturated.

7 0

3 years ago

Read 2 more answers

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago