Write the chemical reaction for hydrogen thiocyanate in water, whose equilibrium constant is Ka. Include the physical states for each species



Write the chemical reaction for thiocyanate ion in water, whose equilibrium constant is Kb. Include the physical states for each species

Answer:

Dear user,

Answer to your query is provided below

When small amount of acid was added to buffered solution, pH will change very less.

Explanation:

Buffer solution resists change in ph on adding small amount of acid or base but when we calculate the value of buffer capacity we take the change in ph when we add acid or base to 1 lit solution of buffer.This contradicts the definition of buffer solution.

Answer:

6 electrons

Explanation:

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Answer:

each p shell can hold max. 6 electorns

i dont know what a q shell is

Covalent compounds are held together with an intra molecular attraction which is weaker than metallic bond

hence covalent compounds exist as liquids, gases and soft solids

Answer:

PF3 : trigonal bipyramidal

Explanation:

PF3 has 4 domains around the central phosphorus (3 shared pairs and one lone pair of electrons), thus the electron geometry that has 4 domains is tetrahedral not  trigonal bipyramidal

From the options the specie that is incorrectly matched is ( 5 ) ; PF₃ : trigonal bipyramidal

The specie PF₃ is composed of 3 shared pairs and one unshared pair of electrons ( i.e. It has 4 domains ) as seen in the  Lewis structure of  PF₃. therefore  when writing its electronic geometry, it should expressed/written as tetrahedral and not trigonal bipyramidal.

Hence we can conclude that The specie that is incorrectly matched is PF3 : trigonal bipyramidal

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Answer:

The 2nd Picture represents the particles in liquids.

Explanation:

Answer:

1. [tex]10.6\; \rm m[/tex] (one decimal place.)

2.[tex]0.79\; \rm g[/tex] (two decimal places.)

3. [tex]16.5\;\rm cm^2[/tex] (three significant figures.)

Explanation:

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: [tex]\rm m[/tex].)

Therefore:

[tex]\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m[/tex]. (Rounded to one decimal place.)

Similarly:

Therefore, the result should be rounded to two decimal places. Its unit should be [tex]\rm g[/tex] (same as the unit of the two inputs.)

[tex]\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g[/tex]. (Rounded to two decimal places.)

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be [tex]\rm cm \cdot cm[/tex], which is occasionally written as [tex]\rm cm^2[/tex].

[tex]\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2[/tex]. (Rounded to three significant figures.)

Answer:Crushed, decreased

Explanation:

Just got it right

Answer:

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

[tex]Q_1=-Q_2[/tex]

Mass of steel= [tex]m_1[/tex]

Specific heat capacity of steel = [tex]c_1=0.452 J/g^oC[/tex]

Initial temperature of the steel = [tex]T_1=2.00^oC[/tex]

Final temperature of the steel = [tex]T_2=T=21.50^oC[/tex]

[tex]Q_1=m_1c_1\times (T-T_1)[/tex]

Mass of water= [tex]m_2= 105 g[/tex]

Specific heat capacity of water=[tex]c_2=4.18 J/g^oC[/tex]

Initial temperature of the water = [tex]T_3=22.00^oC[/tex]

Final temperature of water = [tex]T_2=T=21.50^oC[/tex]

[tex]Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))[/tex]

On substituting all values:

[tex](m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}[/tex]

Answer:

[tex]m_{steel}=24.9g[/tex]

Explanation:

Hello,

In this case, since the water is initially hot, the released heat by it is gained by the steel rod since it is initially cold which in energetic terms is illustrated by:

[tex]\Delta H_{water}=-\Delta H_{steel}[/tex]

That in terms of mass, specific heat and temperature change is:

[tex]m_{water}Cp_{water}(T_f-T_{water})=-m_{steel}Cp_{steel}(T_f-T_{steel})[/tex]

Thus, we simply solve for the mass of the steel rod:

[tex]m_{steel}=\frac{m_{water}Cp_{water}(T_f-T_{water})}{-Cp_{steel}(T_f-T_{steel})} \\\\m_{steel}=\frac{105mL*\frac{1g}{1mL}*4.18\frac{J}{g\°C}*(21.50-22.00)\°C}{-0.452\frac{J}{g\°C}*(21.50-2.00)\°C} \\\\m_{steel}=24.9g[/tex]

Best regards.

Answer:

The new volume is less than the initial volume.

The new volume is 322mL

Explanation:

Based on Boyle's law, the pressure of a gas is inversely proportional to its volume under constant temperature. That means if the pressure of a gas is increased, the volume decrease and vice versa. The formula is:

P₁V₁ = P₂V₂

Where P is pressure and V is volume of 1, initial state and 2, final states.

In the problem, the pressure of the gas increased from 4.04atm to 7.17atm, That means the new volume is less  than initial volume because the gas is compressed occupying less volume.

Replacing in Boyle's equation:

4.04atm*571mL = 7.17atmV₂

Beeing the new volume of the compressed gas 322mL