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Anon25 [30]
2 years ago
12

If two atoms have a difference in electronegativity of 2.0, what type of tond do they have?

Chemistry
1 answer:
stellarik [79]2 years ago
3 0

Answer:

Ionic

Explanation:

got it correct on a quiz :)

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A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter
Ira Lisetskai [31]

Answer:

m H2O = 56 g

Explanation:

  • Q = mCΔT

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

5 0
3 years ago
In which step of the four-stroke engine cycle does the spark plug initiate the combustion reaction?
Dominik [7]
ANSWER
Power stroke

EXPLANATION

At the power stroke stage, the piston is forced to move down by the expanding gases as both intake valve and exhaust valve are closed.
A spark from the spark plug contributes to ignition of the compressed fuel air mixture to released energy to perform work.
4 0
2 years ago
Read 2 more answers
Find the horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40m/s​
vfiekz [6]

To Find :

The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s​.

Solution :

The horizontal range of a projectile is given by :

R = \dfrac{u^2 sin 2\theta}{g} ( Here, g is acceleration due to gravity = 10 m/s² )

Putting all value in above equation :

R = \dfrac{40^2 \times sin (2 \times 15)}{10} \ m\\\\R = \dfrac{1600 \times 1}{2\times 10} \ m\\\\R = 80 \ m

Therefore, the horizontal range of projectile is 80 m.

4 0
3 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
2 years ago
The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the
AysviL [449]

Answer:

4.42x10⁻¹⁹ J/molecule

Explanation:

At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:

E = 614 - 348 = 266 kJ/mol

Knowing that 1 kJ = 1000 J, E = 266,000 J/mol

By Avogadro's number, 1 mol = 6.02x10²³ molecules, thus:

E = 266,000 J/mol * 1mol/6.02x10²³ molecules

E = 4.42x10⁻¹⁹ J/molecule

7 0
3 years ago
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