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2 years ago

A welder has 1.873 × 102 g Fe2O3 and 94.51 g Al in his welding kit. Which reactant will he run out of first?

Chemistry

2 answers:

maksim [4K]2 years ago

5 0

This is the process i used to find the answer

nalin [4]2 years ago

5 0

Answer:

Fe2O3

Explanation:

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Ionic bonds happened between?

Leni [432]

Answer:

$\large \boxed{\sf metal \ and \ non \ metal}$

Explanation:

Ionic bond is a bond between metals and non-metals with strong forces of attraction between charged ions.

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2 years ago

Which sample of ethanol has particles with the highest average kinetic energy?

ale4655 [162]

The second one.
Remember, temperature is the AVERAGE kinetic energy. So, looking for highest average kinetic energy means looking for highest temperature, which is 55.
Note: the amount of the substance does not determine the average kinetic energy.

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2 years ago

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When hydrogen is burned in the presence of oxygen it will form water, as per the equation: 2H2 + O2 -> 2H2O. Which types of r

nadya68 [22]

Answer:

Explanation:

Hydrogen is synthesized to water by adding Oxygen.

Hydrogen is oxidised to water by combustion ( burning in presence of oxygen).

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2 years ago

Energy is just like any other chemical or physical process, where it cannot be created or destroyed. Which Law states this fact?

Ganezh [65]

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Law of Conservation of Energy

Explanation:

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3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

2 years ago