Answer:
5.2 ×10^-4 moles
Explanation:
Equation of reaction:
5C2O4^2- + 2MnO4^- +. 6H^+. ----------> 10CO2. +. 8H2O. + 2Mn^2+
From the information provided in the question:
Volume of potassium permanganate= 34.88ml
Concentration of potassium permanganate= 0.015M
Amount potassium permanganate= 0.015 × 34.88/1000= 5.2 ×10^-4 moles
Answer:
22.44°C will be the final temperature of the water.
Explanation:
Heat lost by tin will be equal to heat gained by the water
![-Q_1=Q_2](https://tex.z-dn.net/?f=-Q_1%3DQ_2)
Mass of tin = ![m_1=18.5 g](https://tex.z-dn.net/?f=m_1%3D18.5%20g)
Specific heat capacity of tin = ![c_1=0.21 J/g^oC](https://tex.z-dn.net/?f=c_1%3D0.21%20J%2Fg%5EoC%20)
Initial temperature of the tin = ![T_1=97.38^oC](https://tex.z-dn.net/?f=T_1%3D97.38%5EoC)
Final temperature =
=T
![Q_1=m_1c_1\times (T-T_1)](https://tex.z-dn.net/?f=Q_1%3Dm_1c_1%5Ctimes%20%28T-T_1%29)
Mass of water= ![m_2=75.7 g](https://tex.z-dn.net/?f=m_2%3D75.7%20g)
Specific heat capacity of water= ![c_2=4.184 J/g^oC](https://tex.z-dn.net/?f=c_2%3D4.184%20J%2Fg%5EoC%20)
Initial temperature of the water = ![T_3=21.52^oC](https://tex.z-dn.net/?f=T_3%3D21.52%5EoC)
Final temperature of water =
=T
![Q_2=m_2c_2\times (T-T_3)](https://tex.z-dn.net/?f=Q_2%3Dm_2c_2%5Ctimes%20%28T-T_3%29)
![-Q_1=Q_2](https://tex.z-dn.net/?f=-Q_1%3DQ_2)
![-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)](https://tex.z-dn.net/?f=-%28m_1c_1%5Ctimes%20%28T-T_1%29%29%3Dm_2c_2%5Ctimes%20%28T-T_3%29)
On substituting all values:
![-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)](https://tex.z-dn.net/?f=-%2818.5%20g%5Ctimes%200.21%20J%2Fg%5EoC%5Ctimes%20%28T-97.38%5EoC%29%29%3D75.7%20g%5Ctimes%204.184%20J%2Fg%5EoC%5Ctimes%20%28T-21.52%20%5EoC%29)
we get, T = 22.44°C
22.44°C will be the final temperature of the water.
2 hydrogen and 1 oxygen makes water
what??? i need more information
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.