Mass, if you know what element you are working with.
Answer:
1. See explanation below
2. Density
3. Masses
Explanation:
1. Your picture is a bit too small to see the values but maybe this will help you.
To determine the maximum maximum mass in grams that triple beam balance can measure all you have to do is add up the maximum of each beam. So all you need to do is see the value at the last notch of each beam.
However, if you are referring to the picture that is attached in the bottom: The answer would be 610g. Because the last notches of each beam are as follows:
100 g
500 g
10 g
So we add that we get 610g.
2. density can be computed using the formula:
D = M/V
where:
D = density
M = mass
V = volume
As you can see in the both figures A and B measure 20 g, this means that their masses are the same. The density of objects can be different when either their masses, or their volumes are different. So even if they have the same mass, they can have different densities because they have different volumes.
3. Force of gravitational attraction between two objects is dependent on the masses of the two objects and the distance. The larger the mass, the stronger the gravitational force of attraction. This means that they have a direct relationship. Now when it comes to distance, the further apart they are the weaker the gravitational force of attraction, or in other words, they are indirectly related.
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)
Work = force*distance
Work = 670 * 5
Work = 3350 Nm
Answer:
The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)
Explanation:
Let's think all the situation.
2 ICl(g) ⇄ I₂(g) + Cl₂(g)
Initially 0.20 - -
Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.
React x x/2 x/2
Because the ratio is 2:1, in the reaction I have the half of moles.
So in equilibrium I will have
(0.20 - x) x/2 x/2
Notice that I have the concentration in equilibrium so:
0.20 - x = 0.060
x = 0.14
So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)
Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).
As we have a volume of 2L, the values must be /2
Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²
Kc = (0.07/2 . 0.07/2) / (0.060/2)²
Kc = 1.225x10⁻³ / 9x10⁻⁴
Kc = 1.36