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kap26 [50]
2 years ago
8

Which reactions are oxidation-reduction reactions? check all that apply. fes(s) 2hcl(aq) → h2s(g) fecl2(g) agno3(aq) nacl(aq)

→ agcl(s) nano3 2c3h6(g) 9o2(g) → 6co2(g) 6h2o(g) fe2o3(s) 3co(g) → 2fe(l) 3co2(g)
Chemistry
1 answer:
Dafna1 [17]2 years ago
7 0

oxidation-reduction reactions are -

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)
  2. Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

For reaction,

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

<u>On reactant side</u>:

Oxidation state of Carbon = +2

Oxidation state of Oxygen = 0

<u>On product side:</u>

Oxidation state of Carbon = +4

Oxidation state of Oxygen = -2

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is oxidizing and the oxygen's oxidation state is decreasing from 0 to -2. As a result, it is decreasing.

For reaction,

                Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

<u>When reacting:</u>

Iron's oxidation state is +3.

Carbon's oxidation state is +2.

<u>On product side:</u>

Iron's oxidation state is zero.

Carbon's oxidation state is +4.

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is being oxidized and the iron's oxidation state is changing from +3 to 0. As a result, it is decreasing.

To learn more about oxidation-reduction from given link

brainly.com/question/5794822

#SPJ4

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Calculate the mass (g) of 1.5 moles of Sulfur. ​
just olya [345]

Answer:

48 g S

Explanation:

Step 1: Define

Molar Mass of Sulfur (S) - 32.07 g/mol

Step 2: Use Dimensional Analysis

1.5 \hspace{3} mol \hspace{3} S(\frac{32.07 \hspace{3} g \hspace{3} S}{1 \hspace{3} mol \hspace{3} S} ) = 48.105 g S

Step 3: Simplify

We have 2 sig figs.

48.105 g S ≈ 48 g S

7 0
3 years ago
A 27.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 ∘C.
Wittaler [7]

Answer:

The final temperature of aluminium ≈ 26.32 °C

Explanation:

<u>Step 1:</u> explain the problem

A 27.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 ∘C.

<u>Step 2:</u> Data given

We will use the formule : Q = mcΔT

with Q = heat transfer ( J)

with m = mass of the substance (g)

with c = specific heat ( J/g °C)

with ΔT = change in temperature ( in °C or K)

mass of aluminium = 27.5g

mass of water = 55.5g

specific heat of aluminium = 0.900J/g °C

specific heat of water = 4.186 J/g °C

initial temperature of aluminium T1= 65.9 °C

initial temperature of water T1 =  22.1 °C

final temperature of water and aluminium = TO BE DETERMINED

<u>Step 3:</u> Calculate the initial temperature

To find the final temperature, we have to use the  following formule:

-(Mass of aluminium) * (caluminium)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-27.5g (0.900)(T2 - 65.9) = 55.5g (4.184j/g °C) (T2- 22.1)

-24.75*(T2-65.9) = 232.212 *(T2-22.1)

-24.75T2 + 1631.025 = 232.212T2 -5131,8852

-256.962 T2 = -6762.9102

T2 = 26.32 °C

The final temperature of aluminium ≈ 26.32 °C

6 0
3 years ago
Which group 2A element has the largest ionic radius?
ivann1987 [24]
The answer to the question is radium (Ra)
5 0
3 years ago
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Would using a strainer be filtration?
katrin2010 [14]

Answer:

To some extent, yes

Explanation:

8 0
3 years ago
A liquid sample, with a density of .0915 g/mL has a mass of 17.7 grams.
Arlecino [84]

The relation between density, mass and volume is

Density = \frac{Mass}{Volume}

1) to calculate = volume

given:

Density = 0.0915 g/ mL

mass = 17.7 grams

formula used:

Density = \frac{Mass}{Volume}

Thus

Volume Volume = \frac{Mass}{Density} = \frac{17.7 g}{0.915 \frac{g}{mL} }  = 193.44 mL

Answer: 193.44 mL

2) the mass of sample will be

mass = density X volume = 0.0915\frac{g}{mL} X 76.5 mL = 6.99 = 7 g

Answer: 7 g

5 0
3 years ago
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