Answer:
Multiply 1.25 by 0.04 and divide the result obtained by 1,000
Explanation:
Given: [1 gram = 0.04 ounce, 1 liter = 1,000 milliliter]
1.25 x 0.04 = 0.05 oz
Therefore, 0.05 per 1,000 milliliter
0.05 ÷ 1,000 = 0.00005 oz
Therefore, the density of the gas is 0.00005 oz/mL
Hope this helps! :)
Answer:
2.5 g
Explanation:
Given data:
Masses of sample = 0.12 g, 1.8 g, 0.562 g
Combine mass of samples = ?
Solution:
When we add or subtract the values the number of significant figures after decimal in result must be equal to the given measurement having less number of decimal places.
0.12 g + 1.8 g + 0.562 g
2.482 g
In given three measurements 1.8 has less number of significant figure after decimal point which is only one digit. Thus the final value must contain one digit after decimal.
we will round of 2.482 g.
2.5 g
because the next digit after 4 is 8 that's why we will round 4 to 5.
From the periodic table:
mass of oxygen = 16 grams
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
Therefore, each 100 grams of CaCO3 contains 3 moles of oxygen
To know the number of oxygen moles in 25.45 grams, we will simply do cross multiplication as follows:
number of oxygen moles = (25.45 x 3) / 100 = 0.7636
Answer:
0.702 /s
Explanation:
Rate constant at 
Rate constant at 
Activation energy, 
Use the following equation to calculate 
Use the following equation to calculate
![\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cfrac%7B%5Cmathrm%7BK%7D_%7B2%7D%7D%7B%5Cmathrm%7B~K%7D_%7B1%7D%7D%3D%5Cfrac%7B%5Cmathrm%7BEa%7D%7D%7B%5Cmathrm%7BR%7D%7D%5Cleft%5B%5Cfrac%7B1%7D%7B%5Cmathrm%7B~T%7D_%7B1%7D%7D-%5Cfrac%7B1%7D%7B%5Cmathrm%7B~T%7D_%7B2%7D%7D%5Cright%5D)
Therefore,
![\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28%5Cfrac%7BK_%7B2%7D%7D%7B3.46%20%5Ctimes%2010%5E%7B-2%7D%20%5Cmathrm%7B~s%7D%5E%7B-1%7D%7D%5Cright%29%20%26%3D%5Cfrac%7B50.2%20%5Ctimes%2010%5E%7B3%7D%20%5Cmathrm%7B~J%7D%20%2F%20%5Cmathrm%7Bmol%7D%7D%7B8.314%20%5Cmathrm%7BJK%7D%5E%7B-1%7D%20%5Cmathrm%7B~mole%7D%5E%7B-1%7D%7D%5Cleft%5B%5Cfrac%7B1%7D%7B298%20%5Cmathrm%7B~K%7D%7D-%5Cfrac%7B1%7D%7B350%20%5Cmathrm%7B~K%7D%7D%5Cright%5D)
![\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28%5Cfrac%7BK_%7B2%7D%7D%7B3.46%20%5Ctimes%2010%5E%7B-2%7D%20%5Cmathrm%7B~s%7D%5E%7B-1%7D%7D%5Cright%29%20%26%3D%5Cfrac%7B50.2%20%5Ctimes%2010%5E%7B3%7D%20%5Cmathrm%7B~J%7D%20%2F%20%5Cmathrm%7Bmol%7D%7D%7B8.314%20%5Cmathrm%7BJK%7D%5E%7B-1%7D%20%5Cmathrm%7B~mole%7D%5E%7B-1%7D%7D%5Cleft%5B%5Cfrac%7B52%20%5Cmathrm%7B~K%7D%7D%7B298%20%5Cmathrm%7B~K%7D%20%5Ctimes%20350%20%5Cmathrm%7B~K%7D%7D%5Cright%5D)
hence, the rate constant at 
is 0.702
