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german
3 years ago
7

S SIO₂ +3C Sic +20o Sic ? mass in grams С c-

Chemistry
1 answer:
seraphim [82]3 years ago
6 0

Answer:

26.74g

Explanation:

The equation of the reaction is;

SIO₂ + 3C --> SiC +2CO

From the balanced equation, the relationship between SiC and C is;

3 mol of C produces 1 mol of SiC

Converting mol to mass using; Mass = moles * Molar mass

Mass of SiC = 1 mol * 40.11 g/mol = 40.11g

This means;

3 mol of C produces 40.11g of SiC

2 mol of C produces xg of SiC

3 = 40.11

2 = x

x = 2 * 40.11 / 3 = 26.74g

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Zinc wire is added to an iron(II) nitrate solution.
Advocard [28]

Answer:

Zn(s) + Fe(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Fe(s)

Explanation:

When metal zinc is added to an iron (II) nitrate solution, we can see the following redox reaction:

Zn(s) + Fe(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Fe(s)

Zinc is oxidized since its oxidation number increases from 0 to +2.

Iron is reduced since its oxidation number decreases from +2 to 0.

3 0
2 years ago
Determine the empirical formula of the following compound if a sample contains 5.28 gsn and 3.37 gf.
earnstyle [38]

Answer:

SnF2

Explanation:

you  divide by the smallest number which is 3.37

4 0
1 year ago
A large room contains moist air at 308C, 102 kPa. The partial pressure of water vapor is 1.5 kPa. Determine
andreev551 [17]

Answer:

you tell me

Explanation:

4 0
2 years ago
A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.
Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0
3 years ago
Please help! you are amazing! also brainliest :)
Ede4ka [16]

im pretty sure its A if i did all my reasearch right.

hope this helps luv. <3

8 0
2 years ago
Read 2 more answers
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