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3 years ago

What does the atomic number represent?

Chemistry

2 answers:

Mrrafil [7]3 years ago

8 0

Answer:

Explanation:

saw5 [17]3 years ago

3 0

C. The number of protons in the atoms nucleus.

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What is the change in boiling point for a 0.615m solution of Mgl2 in water?

ivanzaharov [21]

Answer : The change in boiling point is, $0.94^oC$

Explanation :

Formula used :

$\Delta T_b=i\times K_f\times m$

where,

$\Delta T_b$ = change in boiling point = ?

i = Van't Hoff factor = 3 (for MgI₂ electrolyte)

$K_f$ = boiling point constant for water = $0.51^oC/m$

m = molality = 0.615 m

Now put all the given values in this formula, we get

$\Delta T_b=3\times (0.51^oC/m)\times 0.615m$

$\Delta T_b=0.94^oC$

Therefore, the change in boiling point is, $0.94^oC$

4 0

3 years ago

Using the appropriate starting material, (1) show the synthetic steps of the modified Kiliani-Fischer chain extension synthesis

sashaice [31]

Answer:............

Explanation:

Download doc

3 0

3 years ago

How many liters of O2 at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?

stellarik [79]

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L

7 0

3 years ago

What tool measures mass?

Tatiana [17]

Answer:

balances and scales, measurement transducers, vibrating tube sensors, Newtonian mass measurement devices and the use of gravitational interaction between objects.

Explanation:

4 0

3 years ago

The concentration of carbon monoxide (CO), a common air pollutant, is found in a room to be 5.7 x 10^-3 mg/cm^3. How many grams

damaskus [11]

The amount, in mg, of CO present in the room will be 191,520 mg.

<h3>Stoichiometric problem</h3>

The concentration of the gas in the room is 5.7 x $10^{-3}$ mg/cm3.

The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.

We can obtain the volume of the room as:

350 x 300 x 320 = 33,600,000 cm3

The concentration is in mg/cm3, meaning that it is mass/volume.

Thus:

mass = concentration x volume = 5.7 x $10^{-3}$ mg/cm3 x 33,600,000 cm3

= 191,520 mg

The mass of CO in the room is 191,520 mg

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

5 0

1 year ago