How many bonds are between the two carbon atoms in C2H2?

What is oxidation number of S in H2SO5??

mote1985 [20]

★ « what is oxidation number of S in H2SO5?? » ★

it's 6!!

Explanation:

Oxidation number of S in H2SO5 is 6 .

3 0

2 years ago

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I am in desperate need of help will give 40 points to whoever helps me!!!!! I have a quiz tomorrow but I have no idea how to do

serious [3.7K]

All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.
Freezing point depression or Boiling point elevation:

ΔT = -K (m) (i)

ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent.

K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent.

m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent.

i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.

5 0

3 years ago

A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

6 0

2 years ago

if 28.5 g of calcium hydroxide is dissolved in enough water to make 185g of solution what is the percent by mass of calcium hydr

DENIUS [597]

The percent by mass of calcium hydroxide in the solution : 15.41%

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Mass of solute (Ca(OH₂-Calcium hydroxide) : 28.5

Mass of solution = 185 g

$\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%\\\\\%mass=\dfrac{28.5~g}{185~g}\times 100\%\\\\\%mass=15.41\%$

6 0

2 years ago

How does adding an atom affect the position of existing atoms or lone pairs?

Paha777 [63]

Adding an atom will increase the repulsion between existing atoms and lone pairs. Added atom will result in bond pair-bond pair and bond pair-lone pair repulsion. The magnitude of the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion. The added atom will change the electron geometry and bring about a distortion in the molecular geometry.

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2 years ago

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