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kirza4 [7]
2 years ago
7

Balancing Chemical Equation C=H2=CH4

Chemistry
1 answer:
svlad2 [7]2 years ago
4 0

Answer:

C + 2H2 ⇒ CH4

Explanation:

In order to balance a chemical equation you need to make sure that the number of atoms on both sides are equal

C + H2 = CH4

C = 1

H = 2

Products:

C = 1

H = 4

H2 = 2 × 2 = 4

C + 2H2 ⇒ CH4

Hope this helps.

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Eugene describes the physical property of a material as “sweet and floral.” What physical property of the material is Eugene mos
nataly862011 [7]

Answer:

D

Explanation:

5 0
3 years ago
A military jet cruising at an altitude of 12.0kmand speed of 1900./kmh burns fuel at the rate of 74.4/Lmin. How would you calcul
anygoal [31]

Answer: amount = 2466.95L

Explanation:

given that the speed is = 1900./kmh i.e. 1hr/900km

distance = 1050km

the fuel burns at a rate of 74.4 L/min

therefore the amount of fuel that the jet consumes on a 1050.km becomes;

total fuel used = time × fuel burning rate

where time = distance / speed

∴ total fuel used (consumed) = time × fuel burning rate

total fuel consumed = (1050km × 1hr/1900km) × (60min/ 1hr × 74.4L/1min)

total fuel consumed = 2466.95L

6 0
3 years ago
The number of moles in 4.27×10^22 particles of silver is
bazaltina [42]

Answer:

mol=0.0709mol

Explanation:

Hello,

In this since one mole equals 6.022x10²³ particles of silver by means of the Avogadro's number, we can compute the moles in 4.27x10²² particles as shown below:

mol=4.27x10^{22}particles*\frac{1mol}{6.022x10^{23}particles}\\\\mol=0.0709mol

Best regards.

3 0
3 years ago
PLEASE HELP BEFORE 7 A.M. PACIFIC TIME <br> SEE ATTACHED.
anastassius [24]
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
5 0
3 years ago
Read 2 more answers
A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r
slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

8 0
3 years ago
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