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slava [35]
2 years ago
9

How many grams are 2.50 * 10^23 molecules of CaO?

Chemistry
1 answer:
jolli1 [7]2 years ago
5 0

134.56 grams are 2.50 * 10^23 molecules of CaO.

Explanation:

According to Avagadro hypothesis, 1 mole of a substance contain 6.023 x 10^23 molecules or particles.

Data given:

Number of molecules of CaO =

FORMULA USED IN CALCULATION:

number of molecules = 6.02 x 10^23 x number of moles of CaO

At first number of moles is calculated:

\frac{6.02 x 10^23}{2.50 X 10^23}

= 2.4 moles of CaO is present

to calculate the mass of CaO in grams formula used is

mass = atomic mass x number of moles

atomic mass of CaO = 56.07 grams/mole

mass = 2.4 x 56.07

          = 134.56 grams of calcium oxide is present is 2.50 * 10^23 molecules of CaO.

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En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello
MAXImum [283]

Answer:

a) m_{CH_4}=2630kg

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}

m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol

Con ello, los kilogramos de metano que cuestan 0,45 €:

44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0
2 years ago
What is the mass in grams of 6.022×1023 atoms of mass 16.00 amu?
german
16.00 g -------------- 6.02x10²³ atoms
?? g ----------------- 6.022x10²³ atoms

16.00 x (6.022x10²³) / 6.02x10²³ =

=> 16 g


7 0
2 years ago
Which of the following types of molecules always has a dipole moment? Linear molecules with two identical bonds. Trigonal pyrami
natali 33 [55]

Answer:

Trigonal pyramid molecules (three identical bonds)

Explanation:

In trigonal pyramidal molecule  like molecule of ammonia , the vector some of intra- molecular dipole moment is not zero because the bonds are not symmetrically oriented . In other molecules , bonds are symmetrically oriented in space so the vector sum of all the internal dipole moment  vectors cancel each other to make total dipole moment zero.

8 0
2 years ago
How many ethyne molecules are contained in 84.3 grams of ethyne (C2H2)?
ololo11 [35]
(~26grams/mole) and Avogadros # (6.022x10^23) 84.3grams x 1mole/26grams x 6.022x10^23 molecules/mole = 1.95x10^24 molecules of C2H2
6 0
2 years ago
Read 2 more answers
Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2Mg (s) + O2 (g) →→ 2MgO (s) When 4.50 g of magnes
Klio2033 [76]

Answer:

7.46 g

Explanation:

From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.

The mole ratio is 1:1

mole = mass/molar mass

mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole

0.185 mole Mg will tiled 0.185 MgO

Hence, theoretical yield of MgO in g

mass = mole x molar mass

            0.185 x 40.3 = 7.46 g

6 0
2 years ago
Read 2 more answers
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