Question 3

NEED HELP! SCIENCE HW IS DUE IN 1 HOUR!!!!!!!!

almond37 [142]

Answer:

hair is a physical change.

Cooking is a physical and chemical change.

Ice cream melting in the sun is a physical change.

mark me as the best if this helps you

6 0

3 years ago

If you were to place the black probe of the voltmeter on the silver electrode and the red probe on the Zinc electrode: What volt

podryga [215]

Answer:

When the red test lead is positive (+) and the black test lead negative (-), the meter will register voltage in the normal direction.

Explanation

5 0

3 years ago

Oxidation and reduction reactions (redox) involve the loss and gain of electrons. Half-reactions are a way for us to keep track

jok3333 [9.3K]

Answer:

Electrons are lost during oxidation (LEO)

4 0

3 years ago

Consider four different samples: aqueous LiBr , molten LiBr , aqueous AgBr , and molten AgBr . Current run through each sample p

Charra [1.4K]

Answer:

a) Aqueous LiBr = Hydrogen Gas

b) Aqueous AgBr = solid Ag

c) Molten LiBr = solid Li

c) Molten AgBr = Solid Ag

Explanation:

a) Aqueous LiBr

This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.

b) Aqueous AgBr

This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.

c) Molten LiBr

In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.

c) Molten AgBr

The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.

6 0

3 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago