Identify the type of igneous rock that is formed when magma cools above Earth's surface?

when the sum 4.9965 + 2.11 + 3.887 is calculated, to how many decimal places should the answer be reported?

jarptica [38.1K]

Answer:

the correct answer is four decimal places

3 0

3 years ago

Iron(II) sulfide reacts with hydrochloric acid according to the reaction:

Inessa [10]

The balanced chemical reaction would be:

FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)

We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:

Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl

0.646 - 0.48 = 0.166 mol HCl left

8 0

2 years ago

Read 2 more answers

I am a gas. I am in the noble gas family and row 1. I have only 2 valence electrons I glow red-orange when in an electric field.

belka [17]

The element is Helium

Goodluck!

7 0

2 years ago

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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

2 years ago

A salt bridge is used to provide electrons a path to flow from the electrode of one half-cell to another. TRUE FALSE

kicyunya [14]

It is false. The salt bridge is not a path for electrons, but a path for ions to flow from one half-cell to another. It help to balance the charge between the oxidation and reduction vessels.

8 0

3 years ago