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Helga [31]
3 years ago
5

Define physical change and give an example

Chemistry
2 answers:
Musya8 [376]3 years ago
6 0
Physical isn't so chemical

frozen [14]3 years ago
3 0
Physical changes are the physical properties associated with a specific matter. 

Examples include viscosity, malleability, colour, odor, hardness, etc. 
You might be interested in
. For the reaction 2HNO3 + Mg(OH), → Mg(NO3)2 + 2H20, how many grams of magnesium nitrate are produced from 8.00 mol of nitric a
GREYUIT [131]

Mass of Magnesium nitrate produced : 593.2 g

<h3>Further explanation</h3>

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

2HNO₃ + Mg(OH)₂ → Mg(NO₃)₂ + 2H₂0

mol HNO₃ = 8

From the equation, mol ratio of HNO₃ : Mg(NO₃)₂ = 2 : 1, so mol Mg(NO₃)₂ :

\tt \dfrac{1}{2}\times 8=4~moles

Mass Mg(NO₃)₂(MW=148,3 g/mol) :

\tt mass=mol\times MW\\\\mass=4\times 148,3 g/mol\\\\mass=593.2~g

6 0
2 years ago
How many molecules are in 128g of H2O?
Lelechka [254]

Answer:

4.28x10^24 molecules

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. From the above, we understood that 1mole of H2O also contains 6.02x10^23 molecules.

1mole of H2O = (2x1) + 16 = 2 + 16 = 18g

Now, if 18g of H2O contains 6.02x10^23 molecules,

Then 128g of H2O will contain = (128x 6.02x10^23) /18 = 4.28x10^24 molecules

3 0
2 years ago
What do acid formulas have in common?
maks197457 [2]

the answer is H is the cation

Explanation :

The general formula of an acid is represented as,  in which 'H' is hydrogen cation and 'X' is a non-metal or a poly-atomic anion.

For example : etc.

All the acids produces hydrogen ion,  in an aqueous solution while the base produces hydroxide ion,  in an aqueous solution.

5 0
3 years ago
Read 2 more answers
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
2 years ago
Saturns day is roughly 10 hours. it takes about 30 years for Saturn to return to the same point in space. How do saturns periods
leva [86]
I think its c but not sure
8 0
3 years ago
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