Nitrogen atomic# 7 7protons, 7electrons, 7 neutrons
Oxygen atomic# 8 8protons, 8electrons, 8neutrons
Phosphorous atomic# 15 15protons, 15 electrons, 15 neutrons
Sulfur atomic#16 16protons, 16electrons, 16 neutrons
Answer:
4.1 moles
Explanation:
Applying
PV = nRT................ equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT.............. Equation 2
From the question,
Given: V = 35 L , P = 2.8 atm, T = 15 °C = (15+273) = 288 K, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (35×2.8)/(0.083×288)
n = 4.1 moles
<span>Separate this redox reaction into its component half-reactions.
Cl2 + 2Na ----> 2NaCl
reduction: Cl2 + 2 e- ----> 2Cl-1
oxidation: 2Na ----> 2Na+ & 2 e-
2) Write a balanced overall reaction from these unbalanced half-reactions:
oxidation: Sn ----> Sn^2+ & 2 e-
reduction: 2Ag^+ & 2e- ----> 2Ag
giving us
2Ag^+ & Sn ----> Sn^2+ & 2Ag </span>Steve O <span>· 5 years ago </span><span>
</span>
Answer:
1-Pentene
Explanation:
If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.
For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.
Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.
For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.
Answer:
Explanation:
From the given information:
The concentration of metal ions are:
![[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D%20%5Cdfrac%7B0.003474%20%5C%20M%20%5Ctimes%2020.49%20%5C%20mL%7D%7B10.0%20%5C%20mL%7D)
![[Ca^{2+}]=0.007118 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D0.007118%20%5C%20M)
![[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BMg%5E2%2B%5D%20%3D%20%5Cdfrac%7B0.003474%20%5C%20M%5Ctimes%20%2826.23%20%20-%2020.49%20%29mL%7D%7B10.0%20%5C%20mL%7D)
Mass of Ca²⁺ in 2.00 L urine sample is:
= 0.1598 g
Mass of Ca²⁺ = 159.0 mg
Mass of Mg²⁺ in 2.00 L urine sample is:
= 0.3461 g
Mass of Mg²⁺ = 346.1 mg
