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9966 [12]
4 years ago
5

Describe the difference between intensive and extensive propoerties. Which of the following propoerties are intensive: (a) mass;

(b) density; (c) volume; (d) melting point
Chemistry
1 answer:
STatiana [176]4 years ago
4 0

Answer:

The properties <u>that </u><u>do not vary</u><u> with the variation in the quantity of the material are called as </u><u>intensive property</u>.

While the <u>extensive properties</u><u> are those which </u><u>vary with the variation in the quantity of the material</u>.

Intensive properties:

Mass and volume

Extensive properties:

Density and melting point

Explanation:

The properties <u>that </u><u>do not vary</u><u> with the variation in the quantity of the material are called as </u><u>intensive property</u>.

While the <u>extensive properties</u><u> are those which </u><u>vary with the variation in the quantity of the material</u>.

Intensive properties:

Mass and volume, as with increase in quantity of the material mass and volume increases.

Extensive properties:

Density and melting point, as they remain constant for a object or material

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A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
What temperature should vegetables be cooked at to hold them hot?
shtirl [24]

Answer:

135°F

Fruits and vegetables that are cooked for hot holding must be cooked to a temperature of 135°F (57.2 °C).

Explanation:

Hope it helps you...

(◍•ᴗ•◍)

4 0
2 years ago
A chemical reaction can theoretically produce 137.5 grams of product, but in actuality 112.9 grams are
Butoxors [25]
<h3>Answer:</h3>

82.11%

<h3>Explanation:</h3>

We are given;

  • Theoretical mass of the product is 137.5 g
  • Actual mass of the product is 112.9 g

We are supposed to calculate the percentage yield

  • We need to know how percentage yield is calculated;
  • To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.

Thus;

% yield = (Actual mass ÷ Experimental mass) × 100%

            = (112.9 g ÷ 137.5 g) × 100%

            = 82.11%

Therefore, the percentage yield of the product is 82.11 %

7 0
3 years ago
What nuclide undergoes fission to form molybdenum-103, atomic number 42, tin-131, atomic number 50, and two neutrons?
FrozenT [24]

Answer:

The correct answer is Pu, 234.

Explanation:

In the given case, let us consider the reactant as X. Now the mass number (balanced) on both the sides will be,

Mass of X = Mass of Molybdenum + Mass of Tin + Mass of neutrons

M = 1 * 103 + 1 * 131 + 2 * 0

M = 234

Now the atomic number (balanced) on both the sides,

Atomic number of X = Atomic number of Molybdenum + Atomic number of Tin + Atomic number of neutrons

A = 1*42 + 1*50 + 2*1

A = 94

The atomic number 94 is for the element Plutonium, whose symbol is Pu. Thus, the reactant is 234-Pu.

6 0
3 years ago
The heat of reaction for the combustion of propane is –2,045 kJ. This reaction is: C3H8(g) +502 (g) 3CO2 (g) + 4H2O (g). Determi
Ede4ka [16]

Answer:

\Delta _fH_{C_3H_8}=-102.7kJ/mol

Explanation:

Hello,

In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol

Best regards.

8 0
3 years ago
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