The answer is A, biogeochemical cycles.
Answer:
See explanation and image attached
Explanation:
A bond line structure refers to any structure of a covalent molecule wherein the covalent bonds present in the molecule are represented with a single line for each level of bond order.
The bond-line structure of CH3CH2O(CH2)2CH(CH3)2 has been shown in the image attached. We know that oxygen has a lone pair of electrons and this has been clearly shown also in the image attached.
Answer:
Hello
you're answer should be E.HOCH2CH2OH
hope this answer is correct
Answer: 9.9 grams
Explanation:
To calculate the moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D)
a) moles of ![H_2](https://tex.z-dn.net/?f=H_2)
![\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B8.150g%7D%7B2g%2Fmol%7D%3D4.08moles)
b) moles of ![C_2H_4](https://tex.z-dn.net/?f=C_2H_4)
![\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B9.330g%7D%7B28g%2Fmol%7D%3D0.33moles)
![H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)](https://tex.z-dn.net/?f=H_2%28g%29%2BC_2H_4%28g%29%5Crightarrow%20C_2H_6%28g%29)
According to stoichiometry :
1 mole of
combine with 1 mole of
Thus 0.33 mole of
will combine with =
mole of
Thus
is the limiting reagent as it limits the formation of product.
As 1 mole of
give = 1 mole of ![C_2H_6](https://tex.z-dn.net/?f=C_2H_6)
Thus 0.33 moles of
give =
of ![C_2H_6](https://tex.z-dn.net/?f=C_2H_6)
Mass of ![C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g](https://tex.z-dn.net/?f=C_2H_6%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.33moles%5Ctimes%2030g%2Fmol%3D9.9g)
Thus theoretical yield (g) of
produced by the reaction is 9.9 grams
Answer:
V₂ ≈416.7 mL
Explanation:
This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.
where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.
The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,
- V₁= 600 mL
- T₁= 360 K
- T₂= 250 K
Substitute the values into the formula.
- 600 mL /360 K = V₂ / 250 K
Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.
- 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
- 250 K * 600 mL/360 K = V₂
The units of Kelvin cancel, so we are left with the units of mL.
- 250 * 600 mL/360=V₂
- 416.666666667 mL= V₂
Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.
The volume of the balloon at 250 K is approximately 416.7 milliliters.