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2 years ago

Calling all chemists...please help me! Ive attached a screen shot of the problem

Chemistry

1 answer:

Olin [163]2 years ago

6 0

This is an acid-base reaction where HF is the acid and H2O is the base (it's amphoteric and can be an acid or a base). The products would then H3O+ (the conjugate acid) and F- (the conjugate base). Now, we can simply construct a reaction using the found products and reactants. This acid-base reaction would be HF + H2O <--> H3O+ + F-.

Hope this helps!

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How long is a half life for carbon 14?

babunello [35]

Answer:

half-life of 5,700 ± 40 years

Explanation:

6 0

3 years ago

Read 2 more answers

This graph shows two curves pertaining to a hydrogen s orbital.

fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by <em>'n'</em>.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

3 0

2 years ago

Read 2 more answers

For the reaction, 2SO2(g) + O2(g) <--> 2SO3(g), at 450.0 K the equilibrium constant, Kc, has a value of 4.62. A system wa

Nady [450]

Answer:

To the left.

Explanation:

Step 1: Write the balanced reaction at equilibrium

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

Step 2: Calculate the reaction quotient (Qc)

Qc = [SO₃]² / [SO₂]² × [O₂]

Qc = 0.254² / 0.500² × 0.00855

Qc = 30.2

Step 3: Determine in which direction will proceed the system

Since Qc > Kc, the system will shift to the left to attain the equilibrium.

4 0

3 years ago

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

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

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago

How many moles of oxygen atoms are present in 30.5 grams of hydrogen peroxide (h2o2)?

Pepsi [2]

1.8 moles of oxygen atoms are present in 30.5 grams of hydrogen peroxide.

<u>Explanation:</u>

First we have to convert the given weight of hydrogen peroxide to molar mass of hydrogen peroxide. So for this, we have to divide the given weight with the molecular mass of hydrogen peroxide.

$\text {Molecular mass of } \mathrm{H}_{2} \mathrm{O}_{2}=(2 \times 1)+(2 \times 16)=2+32=34 \mathrm{g}$

So,

$\text {Molar mass of } \mathrm{H}_{2} \mathrm{O}_{2}=\frac{30.5}{34}=0.90 \text { moles of } \mathrm{H}_{2} \mathrm{O}_{2}$

Second step, in this moles, 2 molecules of oxygen are present. Thus 1 mole of Hydrogen peroxide consists of 2 moles of oxygen. Then,

$0.90 moles of $\mathrm{H}_{2} \mathrm{O}_{2}=2 \times 0.90=1.8$ moles of oxygen$

So, 30.5 grams of hydrogen peroxide consists of 1.8 moles of oxygen.

4 0

2 years ago