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Mekhanik [1.2K]
1 year ago
15

The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters

Chemistry
1 answer:
Ivenika [448]1 year ago
7 0

Answer:-  4.15*10^-^2^0\mu L

Solution:- It is a volume unit conversion problem where we are asked to convert the volume from cm^3 to microliters.

We know that:

1cm^3 = 1 mL

1mL=10^-^3L

and, 1L=10^6\mu L

Let's use these conversions factors for the desired conversion using dimensional as:

4.15*10^-^2^3cm^3(\frac{1mL}{1cm^3})(\frac{10^-^3L}{1mL})(\frac{10^6\mu L}{1L})

= 4.15*10^-^2^0\mu L

So, the answer is  4.15*10^-^2^0\mu L .

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If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products
puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

  • <u>For a:</u>

The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}

As, K_{eq}>1, the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0
2 years ago
In what situation do we use a volumetric flask, conical flask, pipette and graduated cylinder? Explain your answer from accuracy
Ad libitum [116K]
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring. 
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
3 0
2 years ago
Calculate the Lattice Energy of KCl(s) given the following data using the Born-Haber cycle: ΔHsublimation K = 79.2 kJ/mol IE1 K
DanielleElmas [232]

Answer:

Explanation:

  The following equation relates to Born-Haber cycle

\triangle H_f = S + \frac{1}{2} B + IE_M - EA_X+ U_L

Where

\triangle H_f is enthalpy of formation

S is enthalpy of sublimation

B is bond enthalpy

IE_M is ionisation enthalpy of metal

EA_X is electron affinity of non metal atom

U_L is lattice energy

Substituting the given values we have

-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L

U_L = - 707 KJ / mol

3 0
1 year ago
If a solution at pH 5 undergoes a 1000-fold increase in [OH-], what is the resulting pH?
lara31 [8.8K]
<span>336*280 i believe... i hope  this helps 

</span>
7 0
2 years ago
How many grams of ca(no3)2 can be produced by reacting excess hno3 with 6.33 g of ca(oh)2?
xeze [42]

Answer:

Amount of Ca(NO3)2 produced = 14.02 g

Explanation:

The given reaction can be depicted as follows:

Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O

Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2

Now, Mass of Ca(OH)2 = 6.33 g

Molar mass of Ca(OH)2 = 74 g/mol

Moles\ Ca(OH)2 = \frac{Mass}{Molar\ Mass} = \frac{6.33 g}{74 g/mol} =0.0855

Based on the reaction stoichiometry:

1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2

Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles

Molar mass of Ca(NO3)2 = 164 g/mol

Mass\ Ca(NO3)2 \ produced = moles*molar\ mass \\= 0.0855\ moles*164\ g/mol = 14.02\  g

6 0
2 years ago
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