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3 years ago

(BRAINLIEST QUEATION)

Chemistry

2 answers:

Svetllana [295]3 years ago

8 0

Answer:

1. Methyl orange indicator: This shows pH in a colouess solution if it changes from colourless to red.

2. Phenolphthalein indicator: This shows pH of a coloured solution if it changes from purple to colorless.

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drek231 [11]3 years ago

3 0

Answer:

what are the options?

Explanation:

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What those this best represent Dalton’s law Charles law

lubasha [3.4K]

Ideal Gas Law

Charles Law is a given volume mass of gas varies directly with the kelvin temperatures when the volume remains constant.

Daltons' Laws is, that at constant temperature, and pressure, the pressure of a mixture of gases that doesn't interact will be the sum of pressures of individual gases.

Hope that helps!!!

7 0

4 years ago

Someone help please I don’t know the answer to the question I meeed help

mihalych1998 [28]

Answer: 143.3

137+115+104+263+98 divided by 5.Which gives us 143.3

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3 years ago

Which of the following are behaviors of a gas that can be explained by the kinetic-molecular theory?

My name is Ann [436]

Answer:

b,c,d

Explanation:

gasses exert pressure, all particles of a gas sample move at the same speed. gas particles can exchange kinetic energy when they collide.

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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

What is the name of the binary molecule H2O2? This compound is also known as hydrogen peroxide and is a properly disinfectant or

Masteriza [31]

Answer:

hydrogen dioxide

Explanation:

3 0

4 years ago