All categories

3 years ago

Una muestra de 1 gramo de un elemento contiene 1,5 x 1022 atomos, cual es su masa molar❓❓

1 answer:

7 0

El número de Avogadro es 6,022 x 10^23, y es el número de átomos que hay en un mol de dicho elemento. A su vez un mol es la cantidad de un elemento cuya masa en gramos coincide con el peso atómico.

Por tanto, 6,022 x 10^23 átomos del elemento tienen una masa en gramos igual a su peso atómico. Hacemos una regla de tres:

1 gramo -------- 1,5 x 10^22 átomos

x ------------------ 6,022 x 10^23 átomos

=> x = 40,1 gramos por mol del elemento.

De modo que su peso atómico es 40,1. Se trata del calcio.

Saludos.

You might be interested in

Draw the orbital notation for the element Boron

Wewaii [24]

Answer:

I can't draw but you could draw 2 electrons in the first orbit and 3 electrons in the second orbit.

Explanation:

6 0

2 years ago

Read 2 more answers

2MG+O2 ...........2MGO

Galina-37 [17]

Mg + 1/2 O2 → MgO

1 mol = 24 g of Mg

X mol = 12 g of Mg

x = 0.5 moles of Mg

Mg :MgO = 1:1 (coefficient from equations using mole ratio)

So

0.5 moles of MgO

1 mol MgO = (24+16) g = 40 g

0.5 moles of MgO = 0.5 × 40

= 20 g of MgO produced

5 0

2 years ago

What volume is occupied by 0.104 mol of helium gas at a pressure of 0.91 atm and a temperature of 314 K ?

Aleksandr-060686 [28]

Answer:

The volume will be "2.95 L".

Explanation:

Given:

n = 0.104

p = 0.91 atm

T = 314 K

Now,

The Volume (V) will be:

= $\frac{nRT}{P}$

By putting the values, we get

= $\frac{0.104\times 0.0821\times 314}{0.91}$

= $\frac{2.6810}{0.91}$

= $2.95 \ L$

7 0

2 years ago

A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water

Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be $(C_6H_{12}O)_n$

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

3 0

2 years ago

A 1.000 gram sample of an alcohol was burned in oxygen to produce 1.913 g of CO2 and 1.174 g of H20. The alcohol contained only

ycow [4]

The formula of alcohol are
1.913 g CO2/44 = 0.043 moles of CO2*12 = 0.043 moles of C 1.174 gram H2O/18 = 0.065 moles of H.
O: 1 - 0.516 g C - 13 g H = 0.354 g O/16 = 0.022 moles of O.
we have 0.043:0.130.;0.022 as C:H:O. Diving by the smallest no. 0.022 we get 1.95:5.9:1 or 2:6:1 as C:H:O.

hope this help

4 0

3 years ago

Read 2 more answers