Which could be the catalyst in a chemical reaction?

What Element is found in column 16, period 2?

Art [367]

The answer is Sulfur.

8 0

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

3 years ago

Copper has a specific heat of 0.385 J/gºC.

Anna71 [15]

Answer:

The final temperature is 348.024°C.

Explanation:

Given data:

Specific heat of copper = 0.385 j/g.°C

Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)

Mass of copper = 62.0 g

Initial temperature T1 = 26.7°C

Final temperature T2 = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q = m.c. ΔT

7670 J = 62.0 g × 0.385 j/g °C ×( T2- 26.7 °C )

7670 J = 23.87 j.°C ×( T2- 26.7 °C )

7670 J / 23.87 j/°C = T2- 26.7 °C

T2- 26.7 °C = 321.324°C

T2 = 321.324°C + 26.7 °C

T2 = 348.024°C

The final temperature is 348.024°C.

6 0

4 years ago

In the reaction FeCl2 + 2NaOH ->Fe(OH)2(s) + 2NaCl, if 6 moles of FeCl2 are added to 6 moles of Na0H, how many moles of NaOH

pogonyaev

Answer : The correct option is, (B) 6 mole

Explanation :

Given moles of $FeCl_2$ = 6 moles

Given moles of $NaOH$ = 6 moles

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$FeCl_2+2NaOH\rightarrow Fe(OH)_2+2NaCl$

From the given balanced reaction, we conclude that

As, 1 moles of $FeCl_2$ react with 2 moles of $NaOH$

So, 6 moles of $FeCl_2$ react with $\frac{2}{1}\times 6=12$ moles of $NaOH$

From this we conclude that, $FeCl_2$ is an excess reagent and $NaOH$ is a limiting reagent because the given moles are less than the required moles and it limits the formation of product.

Thus, the number of moles of NaOH used up in the reaction = Required moles of NaOH - Given moles of NaOH

The number of moles of NaOH used up in the reaction = 12 - 6 = 6 moles

Therefore, the number of moles of NaOH used up in the reaction will be, 60 moles

4 0

3 years ago

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How does urban sprawl affect water consumption?

Vesnalui [34]

Answer:

Urban sprawl is the development in urban areas

Due to which people are migrating from rural areas to urban areas to get the jobs or their income sources due to the opportunities.

So, as population is increasing in the urban areas. The food consumption and water consumption will also increase.

So, this is how urban sprawl affect water consumption.

5 0

3 years ago

Read 2 more answers