This is so basic bro u gotta first move
Answer: 7.693 L
Explanation:
To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:

where,
are initial pressure and volume.
are final pressure and volume.
We are given:

Putting values in above equation, we get:

Thus new volume of the gas is 7.693 L
Answer:
If 1,079.75 Joules of heat are added to 77.75 grams of water, by 3.32 degrees Celsius the temperature of water will increase

Explanation:

Here , q = heat added / removed from the substance
m = mass of the substance taken
= Change in temperature
C = specific heat capacity of the substance
In liquid state the value of C for water is :

Given values :
q = 1079.75 J
m = 77.75 gram
Insert the value of C, m , q in the given equation
on transposing ,



1. 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
2. CH₄ + 2O₂ → CO₂ + 2H₂O
3. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
4. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
5. Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃
1)

2)
CuSO_4+Cu_2Cl_2\neq>

Answer:
C. Arsenic
Explanation:
Each period is terminated by a noble gas with a closed valence shell with electronic configuration ns²np⁶. Since noble gases have completely filled orbitals in the valence shell and are very stable, it becomes very difficult to alter their stable arrangement by the addition or removal of electrons. They they exhibit very low chemical reactivity
Noble gas elements are: Neon, Argon, Krypton, Xenon, Radon, Helium (2 electrons in outer shell, stable).