Cu(NO3)2 --> MM187.5558
NiNO3 *COEF2* --> 120.6983
Explanation:
Filtration serves as a method of separation where solid particles that are suspended in a liquid are isolated by passing the mixture through filter paper's pores. This process ensures that the solid particles accumulate on the filter paper and the liquid flows out through the filter paper's pores.
The ordered sequence of the steps provided is:
- Measure and fold the filter paper.
- Insert the filter paper into the funnel, then position the funnel above the Erlenmeyer flask.
- Let the solid/liquid mixture pass through the filter.
- Rinse the filter paper that holds the mixture with water.
- Measure the weight of the dry filter paper along with the copper.
<span>128 g/mol
Applying Graham's law of effusion, we can utilize the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = effusion rate of gas 1
r2 = effusion rate of gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998.
We can now insert the known values into Graham's equation to find m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
Thus, we find m2 to be 127.992.
Rounding to three significant figures yields 128 g/mol</span>
Response: k = 23045 N/m
Clarification:
To determine the spring constant, one must consider the maximum elastic potential energy that the spring can withstand. The kinetic energy of the vehicle should equal at minimum the elastic potential energy of the spring when it is fully compressed. Hence, we express it as:
(1)
M: mass of the vehicle = 1050 kg
k: spring constant =?
v: car speed = 8 km/h
x: maximum spring compression = 1.5 cm = 0.015m
You need to resolve equation (1) for k. Beforehand, convert the speed v to meters per second:
The spring constant calculates to 23045 N/m
Density is defined as the "mass per unit volume" of an object.
Thus, for an object weighing 100 grams with a volume of 100 milliliters, the density calculates to 100 grams / 100 ml.
When weighing the object, if there is water on the scale's surface, it will contribute additional weight, making the object seem heavier than its actual mass. Consequently, you might mistakenly conclude that the density is GREATER than it truly is.
For instance, if there were 5 ml of water on the scale, with water's density being 1 gram per milliliter (1 g/ml), it would add 5 grams to the object's weight. Using the previous example, the object's mass appears as 105 grams instead of 100 grams. Thus, you would calculate:
density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml
Thus, the effect on density would be to misleadingly suggest it is greater.
I hope this is helpful!
Best of luck
