Answer:
A horizontal line on a speed-time graph represents a constant speed. A sloping line on a speed-time graph represents an acceleration. The sloping line shows that the speed of the object is changing. The object is either speeding up or slowing down.
Answer:
84.8 mL
Explanation:
From the question given above, the following data were obtained:
Mass of CuNO₃ = 3.53 g
Molarity of CuNO₃ = 0.330 M
Volume of solution =?
Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:
Mass of CuNO₃ = 3.53 g
Molar mass of CuNO₃ = 63.5 + 14 + (16×3)
= 63.5 + 14 + 48
= 125.5 g/mol
Mole of CuNO₃ =?
Mole = mass / Molar mass
Mole of CuNO₃ = 3.53 / 125.5
Mole of CuNO₃ = 0.028 moles
Next, we shall determine the volume of the solution. This can be obtained as follow:
Molarity of CuNO₃ = 0.330 M
Mole of CuNO₃ = 0.028 moles
Volume of solution =?
Molarity = mole /Volume
0.330 = 0.028 / Volume
Cross multiply
0.330 × Volume = 0.028
Divide both side by 0.330
Volume = 0.028 / 0.330
Volume = 0.0848 L
Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.0848 L = 0.0848 L × 1000 mL / 1 L
0.0848 L = 84.8 mL
Therefore, the volume of the solution is 84.8 mL.
Answer:
2K +F₂→ 2KF
Explanation:
When we balance an equation, we are trying to ensure that the number of atoms of each element is the same on both sides of the arrow.
On the left side of the arrow, there is 1 K atom and 2 F atoms. On the right, there is 1 K and 1 F atom.
Since the number of K atoms is currently balanced, balance the number of F atoms.
K +F₂→ 2KF
Now, that the number of F atoms is balanced on both sides, check if the number of K atoms are balanced.
<u>Left</u>
K atoms: 1
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The number of K atoms is not balanced.
2K +F₂→ 2KF
<u>Left</u>
K atoms: 2
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The equation is now balanced.
(20/23) times 100 is 87%
now the answer asks for percentage ERROR
100-87 =13
so answer is C
Answer:
Y = 62.5%
Explanation:
Hello there!
In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:
Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:
Best regards!
