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2 years ago

What is the name for the ionic compound (NH4)2C2O4

Chemistry

1 answer:

ddd [48]2 years ago

6 0

Ammonium Oxalate (NH4) 2c204 is the formula number and has a mass of 124.0959

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If a bullet travels at 407.0 m/s, what is its speed in miles per hour? Number click to edit mi/h For the same bullet travelling

kaheart [24]

Answer : The speed in miles per hour is 22 mile/hr.

The speed in yard per min is 26617.8 yard/min

Explanation :

The conversion used for meters to miles is:

$1m=100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}$

The conversion used for second to hour is:

$1s=\frac{1}{60}min\times \frac{1hr}{60min}$

The conversion used for meter per second to mile per hour is:

$1\frac{m}{s}=\frac{100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}}{\frac{1}{60}min\times \frac{1hr}{60min}}$

As we are given the speed of 407.0 meter per second. Now we have to determine the speed in miles per hour.

$1m/s=2.2mile/hr$

So, $407.0m/s=\frac{2.2mile/hr}{1m/s}\times 407.0m/s=895.4mile/hr$

Therefore, the speed in miles per hour is 22 mile/hr.

The conversion used for meter to yard

1m = 1.09 yard

The conversion used for second to hour is:

$1s=\frac{1}{60}min$

The conversion used for meter per second to mile per hour is:

$1m/s=\frac{1.09yard}{frac{1}{60min}}$

$1m/s=65.4yard/min$

As we are given the speed of 407.0 meter per second. Now we have to determine the speed in yards per min

$1m/s=65.4yard/min$

So, $407.0m/s=\frac{65.4yard/min}{1m/s}\times 407.0m/s=26617.8yard/min$

Therefore, the speed in yard per min is 26617.8 yard/min

7 0

2 years ago

A buffer is a solution that is a mixture of either a weak acid and its conjugate base or a weak base and its conjugate acid. Whe

leva [86]

Answer:

a. The conjugate base of an acidic buffer will accept hydrogen protons when a strong acid is added to the solution.

b. An acidic buffer solution is a mixture of a weak acid and its conjugate base.

e. The weak acid of an acidic buffer will donate hydrogen protons when a strong base is added to the solution.

Explanation:

Which of the statements correctly describe the properties of a buffer?

a. The conjugate base of an acidic buffer will accept hydrogen protons when a strong acid is added to the solution. TRUE. The conjugate base neutralizes the excess of hydrogen protons.

b. An acidic buffer solution is a mixture of a weak acid and its conjugate base. TRUE.

c. An acidic buffer solution is a mixture of a weak base and its conjugate acid. FALSE. This is a basic buffer solution.

d. The weak acid of an acidic buffer will accept hydrogen protons when a strong base is added to the solution. FALSE. The weak acid will react with the hydroxyl ions from the added base.

e. The weak acid of an acidic buffer will donate hydrogen protons when a strong base is added to the solution. TRUE. These hydrogen protons will form water.

f. The conjugate base of an acidic buffer will donate hydrogen protons when a strong acid is added to the solution. FALSE. It will accept hydrogen protons.

7 0

2 years ago

Which of the following has higher ionization energy - Be or Ca. Explain your choice in terms of attractive force between protons

Vlad [161]

Answer:

Ca has the greater Ionization Energy because the Trend is I/e decreases as you move down a group. Therefore, Ca has the greater I/e

Explanation:

3 0

2 years ago

Molecules are the smallest units of matter with the characteristic properties of a substance.

Vilka [71]

False, Atoms are the smallest units of matter that display both the chemical and physical properties of it, based on the structure of the atom.

3 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago