Answer:
384.2 K
Explanation:
First we convert 27 °C to K:
- 27 °C + 273.16 = 300.16 K
With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:
Where in this case:
We input the data:
300.16 K * 1600 m³ = T₂ * 1250 m³
And solve for T₂:
T₂ = 384.2 K
Answer:
11.4
Explanation:
Step 1: Given data
- Concentration of the base (Cb): 0.300 M
- Basic dissociation constant (Kb): 1.8 × 10⁻⁵
Step 2: Write the dissociation equation
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
Step 3: Calculate the concentration of OH⁻
We will use the following expression.
![[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%3D%5Csqrt%7BKb%20%5Ctimes%20Cb%20%7D%20%3D%20%5Csqrt%7B1.8%20%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.300%20%7D%20%3D%202.3%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 4: Calculate the pOH
We will use the following expression.
![pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6](https://tex.z-dn.net/?f=pOH%20%3D-log%5BOH%5E%7B-%7D%20%5D%3D%20-log%282.3%20%5Ctimes%2010%5E%7B-3%7D%20M%29%20%3D%202.6)
Step 5: Calculate the pH
We will use the following expression.

Answer:
=> 2.8554 g/mL
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 16.59 g
Volume (v) = 5.81 mL
From our question, we are to determine the density (rho) of the rock.
The formula:

Substitute the values into the formula:

= 2.8554 g/mL
Therefore, the density (rho) of the rock is 2.8554 g/mL.
Answer:
Explanation:
70% (vol/vol) means
cotnaimns 70 %(vol/vol) 70 ml of isoprapnol is there in 100 ml of Rubbing sold alcohol.
if it is 200 ml then obvouly it has the 70*2 =140 ml of isoproanol required.
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