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3 years ago

The number of atoms in 5.78 mol NH4NO3. Show your work.

Chemistry

2 answers:

IRINA_888 [86]3 years ago

6 0

Answer:

3.481 x 10²⁴ atoms

Explanation:

Data Given:

Number of moles of NH₄NO₃ = 5.78 mol

Number of atoms of NH₄NO₃ = ?

Solution:

Formula used

no. of moles = no. of atoms / Avogadro's number

As we have to find no. of atoms

So, we have to rearrange the above equation

no. of atoms = no. of moles x Avogadro's number . . . . . (1)

Where

Avogadro's number = 6.022 x 10²³ atoms

Put values in above Equation 1

no. of atoms = 5.78 mol x 6.022 x 10²³ atoms/mol

no. of atoms = 3.481 x 10²⁴ mol

So,

no. of atoms in 5.78 mole NH₄NO₃ = 3.481 x 10²⁴ atoms

umka21 [38]3 years ago

6 0

Answer:

The answer is 3.48 × 10²⁴

Explanation:

Avogadro's constant/number is used to get the actual number of particles (atoms or ions) in a molecule of a substance.

Avogadro's number is 6.022 × 10²³

To get the actual number of atoms in a substance: avogadro's constant is multiplied by the number of moles of the substance

Hence,

number of atoms = 5.78 × 6.022 × 10²³

number of atoms in 5.78 moles of NH₄NO₃ = 3.48 × 10²⁴

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During a phase change, such as melting or boiling, the kinetic energy __. Highlight correct answer

Pepsi [2]

Answer: During a phase change, such as melting or boiling, the kinetic energy increases.

Explanation:

Kinetic energy is the energy obtained by the molecules of a substance due to their motion.

When phase change such as melting or boiling takes place then it means heat is absorbed by the substance. As a result, their molecules more even more rapidly from one place to another due to which more number of collisions take place.

Hence, a change in phase of substance occurs as kinetic energy increases in melting or boiling process.

Thus, we can conclude that during a phase change, such as melting or boiling, the kinetic energy increases.

4 0

2 years ago

Determine the oxidation state of the metal ion in [co(nh3)5br]2+. express your answer as an integer.

Zanzabum

[Co(NH₃)₅Br]²⁺

Ligands and charges on them,

5 × NH₃ = 5 × 0 = 0

1 × Br⁻¹ = 1 × -1 = -1

Charge on sphere = +2

So, putting values in equation,

Co + (0)₅ - 1 = +2

Co + 0 - 1 = +2

Co - 1 = +2

Co = +2 + 1

Co = +3

Result:
Oxidation state of Co in [Co(NH₃)₅Br]²⁺ is +3.

7 0

2 years ago

Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of

Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

5 0

3 years ago

An object has a mass of 441 g and a volume of 10cm3

expeople1 [14]

Answer:

44.1g/cm3

Explanation:

d=m/v

=441/10

=44.1 g/cm3

3 0

3 years ago

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho

ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= $\frac{59}{760}atm$

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

$n = \frac{PV}{RT}$

$n = \frac{0.078 * 15000}{0.082*290}$

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since numbers of moles = $\frac{mass}{molar mass}$

and mass = density × vollume

Then; we can say ;

number of moles = $\frac{density*volume }{molar mass of ethanol}$

number of moles = $\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}$

number of moles = $\frac{&85}{46.07}$

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0

2 years ago