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3 years ago

7.no . with steps please anyone no spam

Chemistry

1 answer:

gulaghasi [49]3 years ago

6 0

Given mass=1.3g
Atomic mass=65u

Molar mass=65g/mol

Now

$\boxed{\sf No\:of\:moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto No\:of\;Moles=\dfrac{1.3}{65}=0.02mol$

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The term used to describe the rapid release of bubbles from a liquid is _____

aksik [14]

The term used to describe the rapid release of bubbles, or rapid release of a gas from a liquid or a solution is called Effervescence. The bubbling of a solution is due to the escape of a gas which may be from a chemical reaction, as in fermenting liquid, or by coming out of a solution after having been under pressure, as in a carbonated drink. For example; soda, champagne among others.

5 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]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

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

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

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

If an equal number of moles of the weak acid HF and the strong base KOH are added to water, is the resulting solution acidic, ba

s344n2d4d5 [400]

Answer:

Being a weak acid and a strong base, where it is diluted in a neutral medium such as water, the basic medium predominates, almost alkaline pH.

Explanation:

The acidity of the solution, being weak, means that its pH is not so low, therefore it will be easier to reach the values of 7 or 7 where alkalinity or basity is indicated.

6 0

3 years ago

Convert 1.7 Kg to cg

Kipish [7]

To answer this lets first see how much 1 kg is equal to in cg.

1 kg = 100000 cg

Now lets multiply:-

100000 × 1.7 = 170000

So, 1.7 kg = 170000 cg

Hope I helped ya!!!

4 0

3 years ago

Read 2 more answers

You have 2.7 moles of carbon. How many atoms do you have?

Gelneren [198K]

Answer:

1.63 × 10²⁴ atoms.

Explanation:

To calculate the number of atoms (N) contained in 2.7moles of carbon, we multiply the number of moles (n) by Avogadro's number (6.02 × 10²³).

That is, N = n × nA

Where;

N = number of atoms

n = number of moles (mol)

nA = Avogadro's numbe

N = 2.7 × 6.02 × 10²³

N = 16.254 × 10²³

N = 1.63 × 10²⁴ atoms.

Hence, there are 1.63 × 10²⁴ atoms in 2.7moles of Carbon.

4 0

2 years ago