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stepladder [879]
3 years ago
13

What is the ground state electron configuration of a neutral atom of titanium

Chemistry
1 answer:
muminat3 years ago
3 0

Answer:

Explanation:

the ground state electron configuration of ground state gaseous neutral titanium is [Ar]. 3d2. 4s2 and the term symbol is 3F2.

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When Earth is between the Moon and the Sun, the fully lit face of the Moon is seen from Earth. This phase of the Moon is called
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3 years ago
If Zn and H2SO4 undergo a single-displacement reaction, what is the balanced equation?
ankoles [38]

Answer:Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

O2Zn(s) + H2SO4(aq) → 22nH(aq) + SO4(s)

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6 0
2 years ago
Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are
emmasim [6.3K]
When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant. 

Let's do the ICE analysis

      2 NH₃ ⇄ N₂ + 3 H₂
I         0        1.3    1.65
C     +2x       -x      -3x
-------------------------------------
E       0.1        ?        ?

The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 mol
Equilibrium N₂ = 1..3 - 0.05 = 1.25 mol

For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is

K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}

Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.
8 0
3 years ago
As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer
Reika [66]

Answer:

We will need 147.772 mL of KH2PO4 to make this solution

Explanation:

For this case we can give the following equation:

H2PO4 - ⇄ H+ + HPO42-

With following pH- equation:

pH = pKa + log [HPO42-]/[H2PO4-]

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

-0.16 =  log [HPO42-]/[H2PO4-]

10^-0.16 = [HPO42-]/[H2PO4-]

0.6918 = [HPO42-]/[H2PO4-]

Let's say the volume of HPO42-= x  then the volume of H2PO4- will be 250 mL - x

Since both have a concentration of 1M = 1 mol /L

If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]

0.6918 = x / (250 - x)

0.6918*250 - 0.6918x = x

172.95 = 1.6918x

x = 102.228 mL

The volume of HPO42- = 102.228 mL

Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL

To control this we can plug this in the pH equation

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

7.05= 7.21 + log (102.228 / 147.772) = 7.05

We will need 147.772 mL of KH2PO4 to make this solution

3 0
3 years ago
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