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Natasha_Volkova [10]
3 years ago
9

Magnesium hydroxide (Mg(OH)2): g/mol Iron(III) oxide (Fe2O3): g/mol

Chemistry
2 answers:
steposvetlana [31]3 years ago
6 0

Molecular mass of Mg(OH)2

= Atomic mass of Mg + 2(Atomic mass of O) + 2(Atomic mass of H)

= 24 g/mol + 2(16 g/mol) + 2(1 g/mol )

= 58 g/mol

Molecular mass of Fe2O3

= 2(Atomic mass of Fe) + 3(Atomic mass of O)

= 2(56 g/mol) + 3(16 g/mol)

= 160 g/mol

alekssr [168]3 years ago
5 0
Magnesium Hydroxide= 58.33
Iron (III) Oxide= 159.70
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bagirrra123 [75]

Answer:

D

Explanation: An exothermic change is Because it has heat.

4 0
3 years ago
1. What is one of the important roles of tropical rainforests in the earth's environment?
Makovka662 [10]

Answer:

1.D

2.C

3.D

4.D

5.C

6 B

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3 0
3 years ago
What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.
ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_A+p_B+p_C...

Given : p_{total} =total pressure of gases = ?

p_{O_2} = partial pressure of oxygen = 100 mm Hg

p_{N_2} = partial pressure of nitrogen = 573 mm Hg

p_{CO_2} = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

p_{H_2O} = partial pressure of water vapor = 47 torr = 47 mm Hg  (1torr=1 mm Hg)

putting in the values we get:

p_{total}=(100+573+40.28+47)mmHg

p_{total}=760.28mmHg

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

5 0
3 years ago
Which reflects more sunlight-dark soil or white sand?
Gnesinka [82]
I think the answer is white sand (sorry if i am wrong) 
4 0
3 years ago
Read 2 more answers
Please help with #2 and #3
Vaselesa [24]

Answer:

2. V_2=17L

3. V=82.9L

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

\frac{V_2}{T_2}=\frac{V_1}{T_1}

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

V_2=\frac{V_1T_2}{T_1}

Therefore, we plug in the given data to obtain:

V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

V=\frac{nRT}{P}

Therefore, we plug in the data to obtain:

V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L

Best regards!

6 0
3 years ago
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