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2 years ago

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An object will sink in a liquid if the density of the object is greater than that of the liquid. the mass of a sphere is 9.83 g.

if the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density = 13.6 g/cm3).

1 answer:

lord [1]2 years ago

5 0

The volumeof this sphere must be less than 0.7228

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According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by ________ valence electron

Mazyrski [523]

According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by__8__ valence electrons.

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2 years ago

Select the NMR spectrum that corresponds best to p-bromoaniline. (see Hint for the structure.) The selected tab will be highligh

lana [24]

Answer:

The NMR spectrum that corresponds best to p-bromoaniline is the one that is attached in the image below.

Explanation:

For the p-bromoaniline 3 types of hydrogen are observed. The first signal that appears at 3.7 ppm would be from the hydrogens of the NH2 group, the hydrogens in ortho position with respect to the NH2 group give a double at approximately 6.54 ppm, and finally the characteristic 7.21 ppm signal is observed for the hydrogens in meta position with with respect to the NH2 group.

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2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final

krok68 [10]

<u>Answer:</u>

P2 = 778.05 mm Hg = 1.02 atm

<u>Explanation:</u>

We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

where P1 = 724 mm hg

V1 = 3.05 L

T1 = 298 K

P2 = ?

V2 = 2.6 L

T2 = 173 K

Substituting the given values in the equation to get:

$\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}$

P2 = 778.05 mm Hg = 1.02 atm

7 0

2 years ago

What is the 2s in 42,256

MissTica

42,256 = 2,000

42,256 = 200

together they'd be 2,200 (if that's what you needed as well)

5 0

2 years ago

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