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expeople1 [14]
4 years ago
12

Una molécula de O2 esta hecho de​

Chemistry
1 answer:
Airida [17]4 years ago
8 0
Dos átomos de oxígeno
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A sample of nitrogen occupies 10.0 liters at 25°C and 98.7 kPa. What would be the volume at 20°C and 102.7 kPa?
krek1111 [17]
To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the first condition of the nitrogen gas sample, we calculate the number of moles.

n = PV / RT
n = (98.7x 10^3 Pa x 0.01 m^3) / (8.314 Pa m^3/ mol K) x 298.15 K
n = 0.40 mol N2

At the second condition, the number of moles stays the same however pressure and temperature was changed. So, the new volume is calculated as follows:

V = nRT / P
V = 0.40 x 8.314 x 293.15 / 102.7 x 10^3
V = 9.49 x 10^-3 m^3 or 9.49 L
7 0
3 years ago
Identify an alkene and carboxylic acid using primary observations
ehidna [41]

Answer:

The general formula for the carboxylic acids is C nH 2n+1COOH (where n is the number of carbon atoms in the molecule, minus 1).

Explanation:

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4 0
3 years ago
Write 0.000443200 to two significant figures
Tju [1.3M]
0.00044


Zeros to the right of the decimal place are not significant UNLESS they are found in between or after a non-zero number, therefore, we take the 3200 away because those ARE significant so then after you round your answer (if needed) you're left with only two numbers that are significant.
6 0
3 years ago
Calculate the energy E of a sample of 3.50 mol of ideal oxygen gas (O2) molecules at a temperature of 310 K. Assume that the mol
love history [14]

Answer:

13.53 kJ

Explanation:

The energy of a gas can be calculated by the equation:

E = (3/2)*n*R*T

Where n is the number of moles, R is the gas constant (8.314 J/mol.K), and T is the temperature.

E = (3/2)*3.5*8.314*310

E = 13,531.035 J

E = 13.53 kJ

7 0
3 years ago
When 24g of magnesium metal is heated in air, 40g of magnesium oxide is produced. What mass of magnesium would be needed in orde
inysia [295]

Answer: 6.1 g

Explanation:

between Mg and MgO theres a 1;1 MOLE RATIO

here's the balanced equation

2Mg + O2 ==> 2MgO

24g of magnesium is approximately 1 mole of magnesium so it produces 40 g of mgo which is also 1 mole of mgo thus 10/40 =0.25 moles of MgO so 0.25 moles of magnesium would be needed  which is approximately 6.1 g

6 0
3 years ago
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