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Nikitich [7]
3 years ago
15

. A student had a 37.8g sample of compound X. The percent of nitrogen in this compound is 17.2%. How many grams of nitrogen are

present in this sample?
Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Answer:

6.50 grams of the compound X is nitrogen

Explanation:

Step 1: Data given

Mass of compound X = 37.8 grams

Compound contains 17.2 % N

Step 2: Calculate mass of nitrogen

Mass nitrogen = % Nitrogen * total mass

Mass nitrogen = 17.2 % * 37.8 grams = 0.172 * 37.8 grams

Mass nitrogen = 6.50 grams

6.50 grams of the compound X is nitrogen

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What is the mass of an atom with 8 protons, 8 electrons, and 8 neutrons?
mafiozo [28]

Answer:An oxygen atom usually has 8 protons, 8 neutrons, and 8 electrons. Looking at the periodic table, oxygen has atomic number 8 and atomic weight 15.999.

Explanation:

5 0
2 years ago
What is the atomic number of an atom that has fie neutrons and four electrons
Natali [406]

Answer:

4

Explanation:

3 0
3 years ago
Read 2 more answers
Which metal will react spontaneously with Cu2+ (aq) at 25°C?
Gwar [14]

Answer:

Mg  

Explanation:

The standard reduction potentials are

                                              <u>E°/V </u>

Au³⁺(aq ) + 3e⁻  ⟶  Au(s);     1.42

Hg²⁺(aq)  + 2e⁻  ⟶  Hg(l);     0.85

Ag⁺(aq)    +   e⁻  ⟶  Ag(s);    0.80

Cu²⁺(aq)   + 2e⁻ ⟶  Cu(s);   0.34

Mg2+(aq) + 2e- ⟶  Mg(s);   -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

 

5 0
3 years ago
Read 2 more answers
Copper has a specific heat of 0.385 J/gºC.
Anna71 [15]

Answer:

The final temperature is 348.024°C.

Explanation:

Given data:

Specific heat of copper = 0.385 j/g.°C

Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)

Mass of copper = 62.0 g

Initial temperature T1 = 26.7°C

Final temperature T2 = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q = m.c. ΔT

7670 J = 62.0 g × 0.385  j/g °C ×( T2- 26.7 °C )

7670 J = 23.87 j.°C ×( T2- 26.7 °C )

7670 J / 23.87 j/°C = T2- 26.7 °C

T2- 26.7 °C = 321.324°C

T2 = 321.324°C + 26.7 °C

T2 = 348.024°C

The final temperature is 348.024°C.

6 0
3 years ago
A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
3 years ago
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