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2 years ago

Calculate the density of a sample of 1.00 mole of NH3 at 793mmHg and -9.00 C

Chemistry

1 answer:

Andrews [41]2 years ago

5 0

Answer:

A mixture of 2.00 moles of H., 3.00 moles of NH3, 4.00 moles of Co, and 5.00 moles.

Explanation:

A mixture of 2.00 moles of H., 3.00 moles of NH3, 4.00 moles of Co, and 5.00 moles.

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What will happen when a piece of magnesium metal is dropped into a beaker containing a l M solution of copper(1) chloride?

Julli [10]

Answer:

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Explanation:

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7 0

1 year ago

At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)

viktelen [127]

Answer: $K_c$ for this reaction at this temperature is 0.029

Explanation:

Moles of $HBr$ = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of $HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M$

The given balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial conc. 0.500 M 0 M 0 M

At eqm. conc. (0.500-2x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}$

Equilibrium concentration of $[Br_2]$ = x = 0.0955 M

Now put all the given values in this expression, we get :

$K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}$

$K_c=0.029$

Thus $K_c$ for this reaction at this temperature is 0.029

7 0

2 years ago

a) (1 point) Build anthracene, optimize its geometry and examine its structure. Describe its shape. b) (1 point) Measure the C-C

oksano4ka [1.4K]

Answer:

a) The structure of anthracene is planar with all the pi electrons delocalized in the structure to maintain aromaticity.

b) The C-C bond length in anthracene is about 140 pm with all the bond lengths being similar to each other.

The standard C-C bond length is 154 pm while standard C=C bond is about 134 pm. Therefore the bond length in anthracene is smaller than standard C-C bond length and longer than standard C=C bond length. This can be explained from the fact that the C-C bonds in anthracene has be mixed characteristics of single and double bond because of the delocalization of pi electrons over the whole structure. As a result, they are neither fully single nor fully double bond in nature. Hence the observed bond lengths.

c) This molecule is not flat. The N-atom is sp3 hybridized here and the H-atom attached to N will remain out of plane.

Explanation:

8 0

2 years ago

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No

liraira [26]

Answer:

Molality = 0.066 m

Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The freezing-point depression is expressed by:

ΔT=Kf * m

We put the data given by the problem and solve for m:

2.63 °C = 40°C·kg/mol * m

m = 0.06575 m

For the calculation of the molar mass: Molality is defined as moles of solute per kilogram of solvent:

0.06575 m = Moles reserpine / kg camphor

25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We calculate moles of reserpine:

0.06575 m = Moles reserpine / 0.025 kg camphor

Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate the molar mass:

1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.

8 0

2 years ago

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t

lana [24]

yes 2.5 is correct for plato!

8 0

2 years ago