Answer:
5.63 g
Explanation:
Step 1: Write the balanced equation
CuBr₂(aq) + 2 AgCH₃CO₂(aq) ⇒ 2 AgBr(s) + Cu(CH₃CO₂)₂(aq)
Step 2: Calculate the reacting moles of copper (II) bromide
30.0 mL of 0.499 M CuBr₂ react. The reacting moles of CuBr₂ are:
Step 3: Calculate the moles formed of silver (I) bromide
The molar ratio of CuBr₂ to AgBr is 1:2. The moles formed of AgBr are 2/1 × 0.0150 mol = 0.0300 mol.
Step 4: Calculate the mass corresponding to 0.0300 mol of AgBr
The molar mass of AgBr is 187.77 g/mol.
Answer:
Sis I think it happened with me but I am not able to remember if u want u can share if it happened with u
A bond is non polar if it is between same atoms and polar if it is between different atoms.
Same atoms are like two dogs of same strength pulling a bone towards towards each other. But when it’s different atoms it’s like a big dog and small dog then the bone is more towards bigger dog. So it’s the same way in bonds.
Bonds are made up of electrons, when the more stronger pulling atom is present than other the electrons are more towards it and as a result we have polar bond. There is development of a kind of a negative pole and a positive pole.
The stronger atom has electrons towards itself so it has a little more negative charge while the other atom has positive charge. This makes bond polar.
So just look for bond between two different atoms, it would be polar.
Look at the pic below to see the answer.
Marked with green is bond between same atoms... one carbon and another carbon so it is not polar and test marked with blue are polar.
Well the answer should have been 10 but since the bonds at 3 and 8 are two of same type we count only one of them.
The answer is 8... well the answer should be 10 otherwise... discuss it with ur teacher
<u>Answer:</u> The specific heat of metal is 0.821 J/g°C
<u>Explanation:</u>
When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of metal = 30 g
= mass of water = 100 g
= final temperature = 25°C
= initial temperature of metal = 110°C
= initial temperature of water = 20.0°C
= specific heat of metal = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
![30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]](https://tex.z-dn.net/?f=30%5Ctimes%20c_1%5Ctimes%20%2825-110%29%3D-%5B100%5Ctimes%204.186%5Ctimes%20%2825-20%29%5D)
Hence, the specific heat of metal is 0.821 J/g°C
