Answer:
Mass = 381.28 g
Explanation:
Given data:
Number of moles of HNO₃ = 16 mol
Mass of Cu needed to react with 16 mol of HNO₃ = ?
Solution:
Chemical equation:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO
Now we will compare the moles of Cu with HNO₃ from balance chemical equation.
HNO₃ : Cu
8 : 3
16 : 3/8×16 = 6
Mass of Cu needed:
Mass = number of moles × molar mass
Mass = 6 mol × 63.546 g/mol
Mass = 381.28 g
OMG THERE'S SPIDER BEHIND YOU!!! jk XD lets get back to the question.....
example of omnivores would be us humans but since you said ANIMALS then :
BEARS - bears are omnivores they feed on meat like fish and plants like grass or dandelion.
RACCOONS - their omnivores too they feed on meat like rats (ew), fish, frogs..etc they also eat plants like any kind fruit, grains, nuts ( i dont think all kind tho).
so yup those are two examples :D
Answer:- 13.6 L
Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.
Standard pressure is 1 atm that is 101.325 Kpa.
Boyle's law equation is:
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
From given information:-
= 58.7 Kpa
= 23.5 L
= 101.325 Kpa
= ?
Let's plug in the values and solve it for final volume.
![58.7Kpa*23.5L=101.325Kpa*V_2](https://tex.z-dn.net/?f=58.7Kpa%2A23.5L%3D101.325Kpa%2AV_2)
On rearranging the equation for ![V_2](https://tex.z-dn.net/?f=V_2)
![V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7B58.7Kpa%2A23.5L%7D%7B101.325Kpa%7D)
= 13.6 L
So, the volume of hydrogen gas at STP for the given information is 13.6 L.
It’s a 50 50 chance unless one parent has a dominate gene
Answer:
5 L
Explanation:
Given data
- Initial pressure (P₁): 1 atm
- Initial volume (V₁): 2.5 L
- Final pressure (P₂): 0.50 atm
For a gas, there is an inverse relationship between the pressure and the volume. Mathematically, for an ideal gas that undergoes an isothermic change, this is expressed through Boyle's law.
![P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{1atm \times 2.5L}{0.50atm}= 5 L](https://tex.z-dn.net/?f=P_1%20%5Ctimes%20V_1%20%3D%20P_2%20%5Ctimes%20V_2%5C%5CV_2%20%3D%20%5Cfrac%7BP_1%20%5Ctimes%20V_1%7D%7BP_2%7D%20%3D%20%5Cfrac%7B1atm%20%5Ctimes%202.5L%7D%7B0.50atm%7D%3D%205%20L)