What concept explains why some objects heat up faster than others when exposed to the same heat source?

In the equation. C + O2 --------> CO2; CO2 is the

lubasha [3.4K]

Answer: Product

Explanation:

A Chemical reaction shows how substances, written in symbols known as reactants, are converted to different substances called products. An arrow from the left hand side (reactants) points to right hand side ( products )shows the direction of flow of reaction.

In the equation. C + O2 --------> CO2

reactant------> product

CO2 is the product

7 0

3 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

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

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

There are three sets of sketches below, showing the same pure molecular compound (hydrogen chloride, molecular formula ) at thre

RideAnS [48]

Answer:

The correct answer is - option C.

Explanation:

Given: the melting point of HCl is

-114.8 °C, which suggests that below this temperature HCl will be solid.

and, since the boiling point of HCl is - 85.1 °C. It is also suggested that above this temperature HCl will be gas, Therefore.

Solid -114.8 - Ordered arrangement

Liquid -85.1c - Less orderly arranged

Gaseous - Least orderly arranged

Thus, at —90 °C, HCl will be present 'in the liquid state, At — 1 °C, HCl will be present in the gaseous state and at -129 °C, HCl will be present in the solid-state. So, the molecules will be organized in a more orderly manner .

Thus, the correct answer is - option C

8 0

3 years ago

Choose a slideshow or presentation program to organize your research. Your presentation should contain Information collected In

beks73 [17]

Answer:

Do the work

Explanation:

I know

3 0

3 years ago

What effects do catalysts have on chemical reactions?

cupoosta [38]

Answer:

3

Explanation:

8 0

3 years ago