When we combine two simpler substances into a more complex one, it is known as a reaction.

When water freezes into ice, sme of the properties have changed <br> What stays the same

larisa86 [58]

When water freezes into ice, some of the properties have change. What stays the same.

Answer: 1: Ice Melting is a physical change. When liquid water (H2O) freezes into a solid state (ice), it appears changed; However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass.

Have a goood day!

5 0

3 years ago

NEED HELP FAST!!!!!!!!!!!!

erma4kov [3.2K]

Covalent for the first one

8 0

3 years ago

Cooling the concentrated ammonia and hydrochloric acid before carrying out the experiment increased the time taken for the white

Crazy boy [7]

Hope it helps you :)

8 0

3 years ago

Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

b)

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

3 years ago

PLEASE HELP!! I REALLY NEED HELP

Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na): $\frac{22.99}{58.443}$ = .393

2(Cl): $\frac{35.453}{58.443}$ = .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: $\frac{78.196}{138.204}$ = .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: $\frac{12.011}{138.204}$ = .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: $\frac{47.997}{138.204}$ = .347 <u>Step 3</u>: (.347)(100) = 34.7%

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: $\frac{167.535}{231.531}$ = .724 Step 3: (.724)(100)= 72.4%

2: O : $\frac{63.996}{231.531}$ = .276 Step 3: (.276)(100) = 27.6%

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: $\frac{36.033}{92.094}$ = .391 Step 3: (.391)(100) = 39.1%

2: H: $\frac{8.064}{92.094}$ = .088 step 3: (.088)(100) = 8.8%

2: O: $\frac{47.997}{92.094}$ = .521 step 3: (.521)(100) = 52.1%

3 0

3 years ago