Answer:
<h3>A. Dalton recognized that tiny atoms combined to form complex structures. </h3>
I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture.
Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
<em>Q = +6970 kJ</em>
The fraction of the original amount remaining is closest to 1/128
<h3>Determination of the number of half-lives</h3>
- Half-life (t½) = 4 days
- Time (t) = 4 weeks = 4 × 7 = 28 days
- Number of half-lives (n) =?
n = t / t½
n = 28 / 4
n = 7
<h3>How to determine the amount remaining </h3>
- Original amount (N₀) = 100 g
- Number of half-lives (n) = 7
- Amount remaining (N)=?
N = N₀ / 2ⁿ
N = 100 / 2⁷
N = 0.78125 g
<h3>How to determine the fraction remaining </h3>
- Original amount (N₀) = 100 g
- Amount remaining (N)= 0.78125 g
Fraction remaining = N / N₀
Fraction remaining = 0.78125 / 100
Fraction remaining = 1/128
Learn more about half life:
brainly.com/question/26374513
Answer:
The new temperature is 894 K or 621 °C
Explanation:
Step 1: Data given
Initial volume of the container = 2.000L
Initial temperature = 25.0 °C = 298 K
Volume increased to 6.00 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 2.00L
⇒with T1 = the initial temperature = 25 °C = 298 K
⇒with V2 = the increased volume 6.00 L
⇒with T2 = the new temperature
2.00 L / 298 K = 6.00 L / T2
T2 = 894 K = 621 °C
The new temperature is 894 K or 621 °C
<span>This problem uses the
relationship between Ka and the concentrations of the ions. Calculations are as follows:</span>
<span>
</span><span>1.9 x 10-5</span>= x^2 / (0.25 - x)
<span>x is very small and the denominator is approximately equal to 0.25. Thus, x is 2.2 x 10^-3
</span><span>pH = -log (2.2 x 10^-3)</span> = 2.66
