3 L of a gas at 250 K is changed to a temperature of 300 K. What is the final volume?

Consider the reaction below. 2Al2O3 4Al + 3O2 What is the mole ratio of oxygen to aluminum?

DiKsa [7]

The answer is 3/4.

The coefficient next to each reactant represents the amount of moles. The compound for oxygen is O2 and the compound for aluminum is 4. We can see that the number next to O2 is 3 and the number next to aluminum is 4.

8 0

3 years ago

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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

3 years ago

A chemistry student needs of isopropenylbenzene for an experiment. He has available of a w/w solution of isopropenylbenzene in a

Lera25 [3.4K]

Question:

A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer:

The answer to the question is as follows

The mass of solution the student should use is 23.42 g.

Explanation:

To solve the question we note the following

A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution

Therefore we have 10 g of isopropenylbenzene contained in

100 g * 10 g/ 42.7 g = 23.42 g of solution

Available solution = 120 g

Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.

8 0

3 years ago

A person with a body temperature of 37°C holds an ice cube with a temperature of 0°C in a room where the air temperature is 20.°

Arlecino [84]

The answer is (2). Heat always flows down the temperature gradient, from high temperature to low temperature. Therefore, since the person is the warmest, heat flows from the person to both the ice and the air. Additionally, since the air is warmer than the ice, heat flows from the air to the ice.

8 0

3 years ago

An athlete can run 9 kilometers in 1 hour. If the athlete runs at that same average speed for 30 minutes, how far will the athle

ASHA 777 [7]

The answer to the problem is 4.5 kilometers. you can solve this problem by cross multiplying

4 0

2 years ago