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qwelly [4]
3 years ago
11

How many bonds can carbon (C) form

Chemistry
1 answer:
atroni [7]3 years ago
3 0

Answer: four.

Carbon has four valence electrons, so it can achieve a full outer energy level by forming four covalent bonds. When it bonds only with hydrogen, it forms compounds called hydrocarbons. Carbon can form single, double, or triple covalent bonds with other carbon atoms

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Answer: no

Explanation:

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3 years ago
Which characteristic of a substance is considered a chemical property?
satela [25.4K]

Answer:

reactivity

Explanation:

a chemical reacting to another chemical is considered a Chemical property because chemicals react with other chemicals differently

8 0
3 years ago
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Nickel metal will react with CO gas to form a compound called nickel tetracarbonyl (Ni(CO)4), which is a gas at temperatures abo
Bad White [126]

Answer:

The final total pressure in the bulb will be 0.567 atm.

Explanation:

The equation of the reaction is:

Ni + 4CO → Ni(CO)₄

The pressure in the bulb will be the sum of the pressures of each gas (remaining CO and Ni(CO)₄ produced).

The pressure of each gas can be calculated using this equation:

For the gas Ni(CO)₄:

P(Ni(CO)₄) = n * R * T / V

where:

P(Ni(CO)₄) = pressure of Ni(CO)₄

n = number of moles of Ni(CO)₄.

R = gas constant = 0.082 l amt / K mol

T = temperature

V = volume

So we have to find how many moles of Ni(CO)₄ were produced and how many moles of CO remained unreacted.

We can calculate the initial number of moles of CO with the data provided in the problem:

P(CO) = n * R * T / V

solving for n:

P(CO) * V / R * T = n

Replacing with the data:

1.20 atm * 1.50 l / 0.082 (l atm / K mol) * 346K = n

n = 0.06mol.

Now we know how many moles of CO were initially present.

To know how many moles of Ni(CO)₄ were produced, we have to find how many Ni reacted with CO.

Initially, we have 0.5869 g of Ni, which is (0.5869 g * 1 mol/58.69 g) 0.01 mol Ni.

From the chemical equation, we know that 1 mol Ni reacts with 4 mol CO, therefore, 0.01 mol Ni will react with 0.04 mol CO producing 0.01 mol Ni(CO)₄ (see the chemical equation above).

At the end of the reaction, we will have 0.01 mol Ni(CO)₄ and (0.06 mol - 0.04 mol) 0.02 mol CO.

Now we can calculate the pressure of each gas after the reaction:

PNi(CO)₄ = n * R * T / V

PNi(CO)₄ = 0.01 mol * 0.082 (l amt / K mol) * 346K / 1.50 l = 0.189 atm

In the same way for CO:

P(CO) = 0.02 mol * 0.082 (l amt / K mol) * 346K / 1.50 l = 0.189 atm = 0.378 atm

The total pressure (Pt) in the bulb, according to Dalton´s law of partial pressures, is the sum of the pressures of each gas in the mixture:

Pt = PNi(CO)₄ + P(CO) = 0.189 atm + 0.378 atm = <u>0.567 atm.</u>

6 0
4 years ago
In a chemical reaction at constant temperature, the addition of a catalyst: _______________
Tom [10]

Answer:

affects the equilibrium content

4 0
3 years ago
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A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and t
drek231 [11]

Answer:

(a) ΔU = 7.2x10²

(b) W = -5.1x10²

(c) q = 5.2x10²

Explanation:

From the definition of power (p), we have:

p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} (1)

<em>where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s) </em>  

(a) We can calculate the energy (ΔU) using equation (1):

\Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J  

(b) The work is related to pressure and volume by:

\Delta W = -p \Delta V

<em>where p: pressure and ΔV: change in volume = V final - V initial      </em>

\Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J

(c) By the definition of Energy, we can calculate q:

\Delta U = \Delta W + \Delta q

<em>where Δq: is the heat transfer </em>

\Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J    

I hope it helps you!  

6 0
4 years ago
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