Answer:
more than
Explanation:
In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.
<h2>
Entire trip takes 1.22 seconds.</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 0.866 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.866 + 0.5 x 9.81 x 0.866²
s = 3.68 m
Halfway is 3.68 m
Total height = 2 x 3.68 = 7.36 m
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = ?
Displacement, s = 7.36 m
Substituting
s = ut + 0.5 at²
7.36 = 0 x t + 0.5 x 9.81 x t²
t = 1.22 s
Entire trip takes 1.22 seconds.
The appropriate response is Gallium. It is a concoction component with image Ga and nuclear number 31. It is in gathering 13 of the occasional table and subsequently has similitudes to alternate metals of the gathering, aluminum, indium, and thallium.
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
Answer:
0.54
Explanation:
Draw a free body diagram. There are 5 forces on the desk:
Weight force mg pulling down
Applied force 24 N pushing down
Normal force Fn pushing up
Applied force 130 N pushing right
Friction force Fnμ pushing left
Sum of the forces in the y direction:
∑F = ma
Fn − mg − 24 = 0
Fn = mg + 24
Fn = (22)(9.8) + 24
Fn = 240
Sum of the forces in the x direction:
∑F = ma
130 − Fnμ = 0
Fnμ = 130
μ = 130 / Fn
μ = 130 / 240
μ = 0.54
