Answer:
A velocity-time graph shows how velocity changes over time. The sprinter's velocity increases for the first 4 seconds of the race, it remains constant for the next 3 seconds, and it decreases during the last 3 seconds after she crosses the finish line.
Answer:
There are 5 significant digits in 0.23100.
Explanation:
This is because all non-zero digits are considered significant and zeros after decimal points are considered significant.
Answer:
Reactants: K = 2, S = 1, Co = 1, Cl = 2; Products: K = 2, S = 1, Co = 1, Cl = 2
Explanation:
The reaction is a double replacement reaction so the anions (Cl⁻ and S²⁻) switch places.
<u>1</u> K₂S + <u>1</u> CoCl₂ ⇒ <u>2</u> KCl + <u>1 </u>CoS (balanced chemical equation)
On the reactants and products side, K = 2, S = 1, Co = 1, and Cl = 2.
Hope that helps.
The characteristics of the α and β particles allow to find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.
The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.
The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.
Thorium has several isotopes, with different rates and types of emission:
- ²³²Th emits α particles, it is the most abundant 99.9%
- ²³⁴Th emits β particles, exists in small traces.
In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.
Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.
In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
Learn more about radioactive emission here: brainly.com/question/15176980
Answer:
The percent by mass of nitric acid in the mixture is 11.1 %
Explanation:
Step 1: Data given
Mass of HNO3 = 8.89 grams
Volume of KOH = 27.1 mL = 0. 0271 L
Molarity of KOH = 0.581 M
Step 2: The balanced equation
HNO
3 + KOH → KNO
3 + H
2
O
Step 3: Calculate the moles of KOH
Moles of KOH = molarity KOH * volume
Moles KOH = 0.581 M * 0.0271 L
Moles KOH = 0.0157 moles
Step 4: Calculate moles of HNO3
For 1 mol of KOH we need 1 mol of HNO3
For 0.0157 moles of KOH we need 0.0157 moles of HNO3
Step 5: Calculate mass of HNO3
Mass KOH = moles KOH * molar mass KOH
Mass KOH = 0.0157 moles * 63.01 g/mol
Mass KOH = 0.989 grams
Step 6: Calculate mass % HNO3 in sample
mass % = (0.989 grams / 8.89 grams)*100%
mass % = 11.1 %
The percent by mass of nitric acid in the mixture is 11.1 %
