CHEMISTRY COLLEGE

A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Answers

Answer 1

Answer:

Answer: The concentration of KOH solution is 1.215 M

Explanation:

For the given chemical equation:

$2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)$

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\nM_1=1.50M\nV_1=16.2mL\nn_2=1\nM_2=?M\nV_2=40.0mL$

Putting values in above equation, we get:

$2* 1.50* 16.2=1* M_2* 40.0\n\nM_2=(2* 1.50* 16.2)/(1* 40.00)=1.215M$

Hence, the concentration of KOH solution is 1.215 M

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Consider the following reaction at equilibrium: NO2(g) + CO(g) = NO(g) + CO2(g) Suppose the volume of the system is decreased at constant temperature, what change will this cause in the system? A shift to produce more NO A shift to produce more CO A shift to produce more NO2 No shift will occur

Answers

Answer: Option (d) is the correct answer.

Explanation:

According to Le Chaltelier's principle, when there occurs any change in an equilibrium reaction then the equilibrium will shift in a direction that will oppose the change.

This means that when pressure is applied on reactant side with more number of moles then the equilibrium will shift on product side that has less number of moles.

For example, $NO_(2)(g) + CO(g) \rightleftharpoons NO(g) + CO_(2)(g)$

Since here, there are same number of moles on both reactant and product side. So, when volume is decreased at a constant temperature in this system then there will occur no change in the equilibrium state.

Thus, we can conclude that in the given when volume of the system is decreased at constant temperature, then no shift will occur.

A 51.9g sample of iron, which has a specific heat capacity of 0.449·J·g?1°C?1, is put into a calorimeter (see sketch at right) that contains 300.0g of water. The temperature of the water starts off at 19.0°C. When the temperature of the water stops changing it's 20.3°C. The pressure remains constant at 1atm. Calculate the initial temperature of the iron sample. Be sure your answer is rounded to 2 significant digits.

Answers

Answer:

the initial temperature of the iron sample is Ti = 90,36 °C

Explanation:

Assuming the calorimeter has no heat loss to the surroundings:

Q w + Q iron = 0

Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )

Assuming Q= m*c*( T- Tir)

mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0

Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )

Tir = 90.36 °C

Note :

- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C

- We assume no reaction between iron and water

Final answer:

To calculate the initial temperature of the iron sample, use the equation q = m * c * T, where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, and T is the change in temperature which is 90.36 °C

Explanation:

To calculate the initial temperature of the iron sample, we can use the equation:

q = m * c * T

Where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, andT is the change in temperature. In this case, we know the mass of the iron sample, the specific heat capacity of iron, and the change in temperature of the water. By rearranging the equation, we can solve for the initial temperature of the iron sample.

Thus,

Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )

Tir = 90.36 °C

Learn more about Calculating initial temperature here:

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What is my percent yield of titanium (II) oxide if I react 20 grams of titanium (II) oxide in excess water (that means TiS is my limiting reactant) and my actual yield of titanium (II) oxide is 13 g?

Answers

Answer:

your percent yield Is

Explanation:

1.5384615385

Brainliest plz

Consider the reaction of aqueous potassium sulfate with aqueous g silver nitrate based on the solubility rule predict the product likely to be precipitate write a balanced molecular equation describing the reaction.

Answers

Answer:

K₂SO₄(aq) + 2AgNO₃ (aq) → 2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

Explanation:

Our reactants are: K₂SO₄ and AgNO₃

By the solubility rules, we know that sulfates are insoluble when they react to Ag⁺, Pb²⁺, Ca²⁺, Ba²⁺, Sr²⁺, Hg⁺

We also determine, that salts from nitrate are all soluble.

The reaction is:

K₂SO₄(aq) + 2AgNO₃ (aq) → 2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

MEASUREMENT AND MATTER Interconverting temperatures in Celsius and Kelvins The metal osmium becomes superconducting at temperatures below 655.mk Calculate the temperature at which osmium becomes superconducting in degrees Celsius. Be sure your answer has the correct number of significant digits. Ac

Answers

Answer: $-272^0C$ .

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ and $K$

These units of temperature are inter convertible.

We are given:

Temperature of the gas = $655mK=0.655K$ (1mK=0.001 K)

Converting this unit of temperature into $^0C$ by using conversion factor:

$(t-273.15)^0C=tK$

$273.15K=0^0C$

Thus $0.655K=(0.655-273.15)^0C=-272^0C$

Thus the temperature is $-272^0C$ .

1. (a) What name is given to the law describing the relationship between volume and pressure at constant temperature? Write a mathematical expression that describes this relationship. (2 marks)(b) Sketch a graph of the relationship described in part (a).