2. How are UVC rays different from UVA and UVB rays?

I do not understand this question if someone could help please

zzz [600]

$\bold{\huge{\blue {\underline{ Answer}}}}$

Potassium = K3PO4 , K2O, KF, KBr, KNO2 , K2SO4 
Sodium = Na3PO4 , Na2O, NaF, NaBr, NaNO3 , Na2SO4
Magnesium = Mg3(PO4)2 , MgO, MgF2 , MgBr2 , Mg3NO2 , MgSO4 
Aluminium = AlPO4 , Al2O3 , AlF3 , AlBr3 , AlNO2 , Al2(SO4)3
Iron(|||) = FePO4 ,  Fe2O3 , FeF3 , FeBr3 , FeNO2 , Fe2(SO4)3 
Copper(||) = Cu3(PO4)2 , CuO, CuF2 , CuBr2 , Cu3NO2 , CuSO4
Barium = Ba3(PO4)2 , BaO, BaF2 , BaBr2 , Ba3NO2 , BaSO4 .

4 0

2 years ago

In an experiment 10 grams of a substance has a volume of 24 cm. What is the

egoroff_w [7]

Explanation:

density= mass/volume

= 10/24

= 0.41 g/cm³

8 0

3 years ago

Which acid is the best choice to create a buffer with ph= 3.19?

Crazy boy [7]

The best choice is hypochlorous acid nitrous acid (HNO2) because it has the nearest value of pK to the desired pH.
pKa of nitrous acid is 3.34
If we know pKa and pH values, we can calculate the required ratio of conjugate base (NO2⁻) to acid (HNO2) from the following equation:
pH=pKa + log(conc. of base)/( conc. of acid)
3.19=3.34 + log c(NO2⁻)/c(HNO2)
3.19 - 3.34 = log c(NO2⁻)/c(HNO2)
-0.15 = log c(NO2⁻)/c(HNO2)
c(NO2⁻)/c(HNO2) = 10⁰¹⁵ = 1.41

5 0

4 years ago

Electrons is an excited state are more likely to enter into chemical reactions.

Alex17521 [72]

A. True would be the best answer

3 0

3 years ago

5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points

labwork [276]

Answer: (A) $H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) $H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of $H_{3}O^{+}$ in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

$Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)$

$NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)$

Hence, net ionic equation will be as follows.

$H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

$NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)$

The net ionic equation is as follows.

$H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

7 0

3 years ago