Hey there!:
Molar mass:
H2 = 2.01 g/mol ; H2O = 18.01
Given the reaction:
2 H2 + O2 = 2 H2O
2 * (2.01 ) g H2 ------------- 2 * ( 18.01 ) g H2O
mass H2 --------------------- 1.80 g H2O
mass H2 = 1.80 * 2 * 2.01 / 2* 18.01
mass H2 = 7.236 / 36.02
mass H2 = 0.2008 g
Hope that helps!
Answer:
k = -0.09165 years^(-1)
Explanation:
The exponential decay model of a radioactive isotope is generally given as;
A(t) = A_o(e^(kt))
Where;
A_o is quantity of isotope before decay, k is decay constant and A(t) is quantity after t years
We are given;
A_o = 5 kg
A(10) = 2kg
t = 10 years
Thus;
A(10) = 2 = 5(e^(10k))
Thus;
2 = 5(e^(10k))
2/5 = (e^(10k))
0.4 = (e^(10k))
In 0.4 = 10k
-0.9164 = 10k
k = -0.9164/10
k = -0.09165 years^(-1)
Answer:
A. because there both high pitched voice and low pitched ( girls have sweet, shrill ,high pitched good voice ) where as boys have horase ,low pitched medium voice )
The amount of oxygen that are produced when 1.06 grams of potassium chlorate decompose completely is 0.64 grams.
<h3>What is the relation between mass & moles?</h3>
Relation between the mass and moles of any substance will be represented as:
- n = W/M, where
- W = given mass
- M = molar mass
Moles of potassium chlorate = 1.66g / 122.5g/mol = 0.0135mole
Given chemical reaction is:
2KClO₃ → 2KCl + 3O₂
From the stoichiometry of the reaction, it is clear that:
2 moles of KClO₃ = produces 3 moles of O₂
0.0135 moles of KClO₃ = produces (3/2)(0.0135)=0.02 moles of O₂
Mass of oxygen = (0.02mol)(32g/mol) = 0.64 g
Hence produced mass of oxygen is 0.64 grams.
To now more about mass & moles, visit the below link:
brainly.com/question/18983376
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Answer: The correct option is:

Explanation:
In balanced chemical equation , mass of the reactants is always equal to the mass of the products.

This is the balanced equation because on both sides number of sodium and bromine atoms are same.
On the reactant side there are two sodium atoms and two bromine atoms
On the product side there are also two sodium atoms and two bromine atoms in NaBr.
Mass of reactants (here 2Na and
)= Mass of products(here 2NaBr)
2(23u) + 2(79.9u) = 2(23u + 79.9u) = 205 u