Hydrogen and oxygen bond because when more molecules are present, as is the case with liquid water, more bonds are possible because the oxygen of one water molecule has two lone pairs of electrons, each of which can form a hydrogen bond with a hydrogen on another water molecule.
Answer:
The balanced equations are as follows:
✅ FeCl₃ + 3NH₄OH ➡ Fe(OH)₃ + 3NH₄Cl
✅ CH₄ + O₂ ➡ CO₂ + 2H₂O
✅ 2Fe + 3H₂SO₂ ➡ Fe₂(SO₂)₃ + 3H₂
✅ PCl₅ + 4H₂O ➡ 5HCl + H₃PO₄
Explanation:
In the equation, FeCl₃ + 3NH₄OH ➡ Fe(OH)₃ + 3NH₄Cl, Charlie will adjust the number of atoms in NH₄OH in order for NH₄, Cl and OH to balance in the product side. 3 was placed in front of NH₄OH and NH₄Cl in order the balance it.
In the equation, CH₄ + O₂ ➡ CO₂ + 2H₂O, Charlie will adjust the number of atoms in H₂O. 2 was placed in front of H₂O in order to balance the atoms in the reactant and product sides.
In the equation, 2Fe + 3H₂SO₂ ➡ Fe₂(SO₂)₃ + 3H₂, Charlie will adjust the number of atoms in the reactant side in order for the product side to be balanced.
In the equation, PCl₅ + 4H₂O ➡ 5HCl + H₃PO₄, Charlie will adjust the number of atoms in H₂O and HCl for the equation to be balanced. 4 and 5 were placed in front of H₂O and HCl respectively in order for the equation to be balanced.
In chemical equations, it is required that the number of atoms of each element found in the reactants must be equal to the number of atoms of each element found in the products. This makes the chemical equation balanced.
Answer:![\dfrac{1}{60} ms^{-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B60%7D%20ms%5E%7B-1%7D)
Explanation:
Speed of an object is defined as the ratio of the distance covered by the object to the time taken to cover that distance.
Let
be the speed of the object.
Let
be the distance travelled by the object.
Let
be the time taken by the object.
So,![s=\frac{D}{t}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7BD%7D%7Bt%7D)
s=![\frac{2}{120}=\frac{1}{60}ms^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B120%7D%3D%5Cfrac%7B1%7D%7B60%7Dms%5E%7B-1%7D)
So,the speed of the car is ![\frac{1}{60}ms^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B60%7Dms%5E%7B-1%7D)
if im not mistaken plant cells go through a stage called photosynthesis meaning the cells can create it's own energy (food) off of water, Co2 and sunlight.
Answer:
(a). The work done is 7001 MeV.
(b). The momentum of this proton is
.
Explanation:
Given that,
Speed = 0.993 c
We need to calculate the work done
Using work energy theorem
The work done is equal to the kinetic energy relative to the proton
![W=K.E](https://tex.z-dn.net/?f=W%3DK.E)
![W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7Bm_%7Bp%7Dc%5E2%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%7D-m_%7Bp%7Dc%5E2)
Put the value into the formula
![W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2%7D%7B%5Csqrt%7B1-%28%5Cdfrac%7B0.993c%7D%7Bc%7D%29%5E2%7D%7D-1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2)
![W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2%7D%7B%5Csqrt%7B1-%280.993%29%5E2%7D%7D-1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2)
![W=1.122\times10^{-9}\ J](https://tex.z-dn.net/?f=W%3D1.122%5Ctimes10%5E%7B-9%7D%5C%20J)
![W=7001\ MeV](https://tex.z-dn.net/?f=W%3D7001%5C%20MeV)
(b). We need to calculate the momentum of this proton
Using formula of momentum
![p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7Bm_%7B0%7Dv%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%7D)
Put the value into the formula
![p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes0.993c%7D%7B%5Csqrt%7B1-%28%5Cdfrac%7B0.993c%7D%7Bc%7D%29%5E2%7D%7D)
![p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes0.993c%7D%7B%5Csqrt%7B1-%280.993%29%5E2%7D%7D)
![p=1.404\times10^{-26}c](https://tex.z-dn.net/?f=p%3D1.404%5Ctimes10%5E%7B-26%7Dc)
![p=4.20\times10^{8}\ kg-m/s](https://tex.z-dn.net/?f=p%3D4.20%5Ctimes10%5E%7B8%7D%5C%20kg-m%2Fs)
Hence, (a). The work done is 7001 MeV.
(b). The momentum of this proton is
.