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QveST [7]
4 years ago
10

How many milliliters is 0.55 L? O A. 1000 ml OB. 0.00055 mL O C. 550 ml D. 55 mL

Chemistry
1 answer:
kolbaska11 [484]4 years ago
3 0

Answer:

The Answer is C. 550 milliliters

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During reflux and distillation, how is it possible to lose the product?
Alla [95]
What is reflux? isn't it like when u aren't able to digest something?
5 0
4 years ago
Starting with 250 mL of a 0.250 M solution of HBr; a) Calculate the initial pH of the solution.b) Calculate the pH after adding
maxonik [38]

Answer:

a.

pH =  0.602

b.

pH = 1.5

c.

pH = 7

d.

pH = 12.1

Explanation:

a ) To calculate the pH, use the following equation:

pH = -log [H+]

Hbr is a strong acid, so the [H+] concentration can be calculated as follow:

[HBr] = 0.250 M

As acid Hbr:

Hbr = H+ + Br-

As strong acid HBr dissociates at all, so

[HBr] = [H+] = 0.250 M

So the pH:

<u>pH = -log [0.250 M] = 0.602</u>

<u></u>

<u>b) Calculate the pH after adding 250 mL of 0.125M NaOH</u>

<u></u>

<u>I</u>n this point, the reactions starts:

<u></u>

HBr + NaOH = H2O + NaBr

- First, we gonna find the mol of each reactant:

HBr:

mol = [M] × L

mol = 0.250 M × 0.250 L

mol = 0.0625 mol HBr

NaOH:

mol = [M] × L

mol = 0.125 M × 0.250L

mol = 0.03125 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.03125 mol HBr = 0.03125 mol HBr

These are the moles free in the solution, and we going to use them to calculate the pH:

pH = -log [ H]

pH = - log [ 0.03125 ] = 1.5

<u>c) Calculate the pH after adding 500 mL of 0.125M NaOH</u>

NaOH:

mol = [M] × L

mol = 0.125 M × 0.500L

mol = 0.0625 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.0625 mol HBr = 0 HBr

pH = 7

<u>d) Calculate the pH after adding 600 mL of 0.125M NaOH.</u>

<u>NaOH:</u>

mol = [M] × L

mol = 0.125 M × 0.600L

mol = 0.075 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.075 mol NaOH – 0.0625 mol NaOH = 00125 NaOH

These are the moles free in the solution, and we going to use them to calculate the pH:

pOH = -log [ OH]

pOH = - log [ 0.0125 ] = 1.90

pH + pOH = 14

pH = 14- pOH = 14 – 1.90 = 12.1

4 0
3 years ago
A region of metamorphic rock lies far beneath Earth’s surface. In time, which two ways can the rock make its way upward to Earth
NeX [460]

It can uplift slowly due to the pressure of Earth’s plates.

It can melt into magma and recrystallize as extrusive igneous rock.

8 0
4 years ago
Read 2 more answers
The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles o
cestrela7 [59]

Answer:

The number of moles of salt in one tablespoon is = <u>0.11 mole</u>

<u>Grams </u>cancel each other.

Explanation:

<u>Moles</u> : It is the unit of quantity . It is the mass of the substance present in exactly 12g of C-12.

moles=\frac{given\ mass}{Molar\ mass}

<u>Moles Calculation:</u>

Given mass = 6.37  gram

Molar mass = 58.5 g/mol

moles=\frac{6.37}{58.5}

= 0.1088

= 0.11 mole

<u>Units calculation</u>

moles=\frac{given\ mass}{Molar\ mass}

moles=\frac{g}{g/mole}

moles=\frac{g}{g}\times mole

<u>g ang g cancels each other </u>

moles = moles

<u>Hence unit = gram (g ) cancel each other.</u>

3 0
3 years ago
Read 2 more answers
A compound that is usually used as a fertilizer can also be used as a powerful explosive. the compound has the composition 35.00
maxonik [38]

Answer:

             Empirical Formula  =  NH₄NO₃ (Ammonium Nitrate)

Solution:

Step 1: Calculate Moles of each Element;

                      Moles of N  =  %N ÷ At.Mass of N

                      Moles of N  = 35.0 ÷ 14

                      Moles of N  =  2.5 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 59.96 ÷ 16

                      Moles of O  =  3.7475 mol


                      Moles of H  =  [100% - (%N + %O)] ÷ At.Mass of H

                      Moles of H  = [100% - (35.0 + 59.96)] ÷ 1.008

                      Moles of H  = [100% - 94.96] ÷ 1.008

                      Moles of H  = 5.04 ÷ 1.008

                      Moles of H  =  5 mol

Step 2: Find out mole ratio and simplify it;

                N                                        H                                     O

               2.5                                       5                                3.7475

            2.5/2.5                                5/2.5                          3.7475/2.5

                 1                                         2                                     1.5

Multiply Mole Ratio by 2,

                 2                                         4                                     3

Result:

         Empirical Formula  =  N₂H₄O₃

Or,

         Empirical Formula  =  NH₄NO₃

This empirical formula is also a Molecular Formula for Ammonium Nitrate a well known Fertilizer and often misused in the formation of Explosives.

4 0
3 years ago
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