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8_murik_8 [283]
2 years ago
13

Galena, a mineral of lead, is a compound of the metal with sulfur. Analysis shows that a 2.34-g sample of galena contains 2.03 g

of lead. Calculate the:
(a) mass of sulfur in the sample;
Chemistry
1 answer:
dolphi86 [110]2 years ago
7 0

The mass of sulfur in the sample is 0.31g

To calculate mass of sulfur in sample: given that amount galena is 2.34g and mass of lead is 2.03 g as galena contain both lead and sulfur the mass of sulfur = mass of galena- mass of lead , that is, 2.34- 2.03=0.31g.The molecular mass is a algebraic sum of atomic mass of each element in the compound thus the mass of the compound can be obtained by adding atomic of each element. Lead(II) sulphide occurs naturally as the mineral galena, often known as lead glance (PbS). It is a significant source of silver and the most significant lead ore.

To learn more about mass:

brainly.com/question/19694949

#SPJ4

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Why does ice float in water?
MaRussiya [10]
Ice has a lower density than the density of water.
5 0
3 years ago
Which of the following intermolecular forces is the strongest?
Tanzania [10]

Answer:

A. Hydrogen Bonding

Explanation:

6 0
3 years ago
If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years
Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually, radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g

8 0
3 years ago
Can someone help me? It needs to have a diagram that has arrows.
daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)

The expression for enthalpy change is,

\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]

Putting the values we get :

\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]

-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]

H_f_{C_4H_{10}=-125kJ/mol

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0
3 years ago
A galvanic^cell consists of left ompartmcnt with a tin elcctrode in contact with 0.1 M Sn(NO_3)_2(aq) and a right compartment wi
lana66690 [7]

Answer:

Electrons will flow from left to right through the wire.

Pb^2+ ions will be reduccd to Pb metal.

The concentration of Sn2+ ions in the left compartment will increase.

Explanation:

Looking at the relative electrode potentials of the two metals

Sn= -0.14

Pb=-0.13

Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.

6 0
2 years ago
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