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3 years ago

Please help!!

Chemistry

1 answer:

maria [59]3 years ago

5 0

Answer:

Transition metals!

Explanation:

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Bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter

Alisiya [41]

Answer:Hence, the bond length in HCl is 125 pm.

Explanation:

Bond length : It is an average distance between the nuclei of two bonded atoms in a molecule.

Also given that bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter-nuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Beyond this if atoms come closer to each other then their will be repulsion between them.

So, the bond length between the Hydrogen and Chlorine atom in HCl molecule is :

$Bond length=radius _{hydrogen}+radius_{chlorine}=25pm+100pm=125pm$

Hence, the bond length in HCl is 125 pm.

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3 years ago

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Roberto and his dad drove from town A to town B taking the long way, ate lunch and then came back using a

alina1380 [7]

Answer:c which is 70

Explanation:

i took the test and got it right

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3 years ago

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Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

qaws [65]

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J$

$E_{ionization}=5.45\times 10^{-19}\ J$

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3 years ago

What is the mass of a rectangular cube with dimensions of 3 cm x 2 cm x to 10 cm and a density of 10 g/cm^3

marta [7]

d=m/V so m=Vd

m=60cm^3 * 10g/cm^3=600g

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3 years ago

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A 250.0-ml sample of ammonia, nh3 (g), exerts a pressure of 833 torr at 42.4 °c. what mass of ammonia is in the container

Lelu [443]

To solve this, let's assume ideal gas behavior.

PV=nRT
Let's solve for n. Convert units to SI units first.

Pressure = 833 torr(101325 Pa/760 torr) = 111,057.53 Pa
Volume = 250 mL(1 L/1000 mL)(1 m³/1000 L) = 2.5×10⁻⁴ m³
Temperature = 42.4 + 273 = 315.4 K

n = (8,314 J/mol·K)(315.4 K)/(111057.53 Pa)(2.5×10⁻⁴ m³)
n = 94.45 mol

The molar mass of ammonia is 17.031 g/mol.
Mass = 94.45*17.031 = <em>1,608.51 g ammonia</em>

6 0

4 years ago