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3 years ago

Salts usually have a low melting point true or false

1 answer:

3 0

Hello!

This is false!! It is the other way around. They usually have a very high melting point. For example table salt is around 800deg C which is approx just under 1500deg F.

Hope this helps. Thank you

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How is copper that is suitable for making electrical wires produced?

patriot [66]

Copper is a good conductor of heat and electricity so it will not fry easily

4 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

2 years ago

How many moles of sulfur dioxide are in 2.26x10^33 sulfur dioxide molecules?

Alexus [3.1K]

Answer:moles = no. of molecules / Avogadro's number

= 2.26 x 10^33 / 6.022 x 10^23

= 3752906011

Round to significant figures which is 3 = 3.75 x 10^9 mol

Explanation:

The formula for finding how many moles of a substance when given the amount of molecules is: moles = number of molecules / Avogadro's number

3 0

3 years ago

Read 2 more answers

Hi guys, my question is:

guapka [62]

Answer:

See Explanation Below

Explanation:

A) The rate law can only be on the reactant side and you can only determine it after you get the net ionic equation because of spectators cancelling out. So in this case the rate law is k=[CH3Br]^1 [OH-]^1. The powers are there because the rxn is first order.

B) Since the rxn is first order anything you do to it will be the exact same "counter rxn" per say so since you are decreasing the OH- by 5 the rate will decease by 5

C) The rate will increase by 4 since you are doubling both you have to multiply them both.

8 0

2 years ago

Name two uses for seismometers

NNADVOKAT [17]

Earthquakes and volcanic eruption

8 0

2 years ago