Answer:
The chemist would require to use 43.43 grams.
Explanation:
In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:
- 0.550 mol * 78.96 g/mol = 43.43 g
The chemist would require to use 43.43 grams.
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
Answer:
Carbon dioxide is moving out of the living things.
Explanation:
The food materials eaten by living things contain carbon in the form of complex organic matter. When living things feed, they ingest this complex organic material into their bodies.
During the process of digestion, this complex organic material is broken down to give glucose. Glucose is the energy molecule in living things. Excess glucose in the body is stored as glycogen.
During cellular respiration, glucose is broken down to release carbon dioxide. Hence, at night when the giraffe has stopped eating, cellular respiration continues to occur and carbon dioxide is released, that is, carbon dioxide continues to move out of living things at night.
2 ICl + H2 ----> I2 + 2 HCl
as given that rate is first order with respect to ICl and second order with respect to H2
The rate law will be
Rate = K [ICl] [ H2]^2
b) Given that K = 2.01 M^-2 s^-1
Concentrations are
[ICl] = 0.273 m and [H2] = 0.217 m
Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s
Extrusive Rocks. Igneous rocks which form by the crystallization of magma at the surface of the Earth are called extrusive rocks. They are characterized by fine-grained textures because their rapid cooling at or near the surface did not provide enough time for large crystals to grow.
