<u>Given: </u>
Radius of culvert, r = 0.5 m
Tangential acceleration of the truck, a = 3 m/s2
<u>To determine:</u>
The angular acceleration, α
<u>Explanation:</u>
The tangential acceleration is related to the angular acceleration through the radius as:
a = rα
α = a/r = 3 ms⁻²/0.5 m = 6 s⁻²
Ans: The angular acceleration is 6 s⁻²
Answer:
A. fluorine, 1.79 moles
Explanation:
Given parameters:
Mass of carbon = 87.7g
Mass of fluorine gas = 136g
Unknown:
The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?
Solution:
Equation of the reaction:
C + 2F₂ → CF₄
let us find the number of the moles the given species;
Number of moles = 
C; molar mass = 12;
Number of moles = 
= 7.31moles
F; molar mass = 2(19) = 38g/mol
Number of moles = 
= 3.58moles
So;
From the give reaction:
1 mole of C requires 2 moles of F₂
7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles
But we have 3.58 moles of the F₂;
Therefore, the reactant in short supply is F₂ and it is the limiting reactant;
So;
2 moles of F₂ will produce mole of CF₄
3.58 moles of F₂ will then produce 
= 1.79moles of CF₄
7<span> to 49 10 to 100. 30 Secs. 3. What is the </span>pH<span> value of pure </span>water<span>? 0 3 </span>7<span> 10 ... How do acids </span>taste<span>? </span>bitter sour<span> sweet salty. 30 Secs. </span>7<span>. How do </span>bases taste<span>? </span>bitter<span> ... 8. Which kind of solution would react with a metal? acidic basic </span>neutral water<span> ... cocoa </span>has<span> a </span>bitter taste<span>. It is most likely which of the following? acid </span><span>base neutral</span>
Answer:
The mass of
4.6
×
10
24
atoms of silver is approximately 820 g.
Explanation:
In order to determine the mass of a given number of atoms of an element, identify the equalities between moles of the element and atoms of the element, and between moles of the element and its molar mass.
1
mole atoms Ag=6.022xx10
23
atoms Ag
Molar mass of Ag =#"107.87 g/mol"#
Multiply the given atoms of silver by
1
mol Ag
6.022
×
23
atoms Ag
. Then multiply times the molar mass of silver.
4.6
×
10
24
atoms Ag
×
1
mol Ag
6.022
×
10
23
atoms Ag
×
107.87
g Ag
1
mol Ag
=
820 g Ag
The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is
2.64 x 10⁻⁵ M.
