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3 years ago

Pt2 science...... Select correct answer ♡!

Chemistry

1 answer:

Dafna1 [17]3 years ago

6 0

E should be correct.

The color may change, precipitation may occur, bubbles may form, gas forms, and the temperature changes.

Mark if correct please!

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A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by measuring out

Genrish500 [490]

Answer:

1.26 × 10^-8 M

Explanation:

We are given;

Number of moles of mercury (i) chloride as 0.000126 μmol

Volume is 100 mL

We are required to calculate the concentration of the solution.

We need to know that;

Concentration is also known as molarity is given by;

Molarity = Number of moles ÷ Volume

Number of moles = 1.26 × 10^-10 Moles

Volume = 0.01 L

Therefore;

Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L

= 1.26 × 10^-8 M

Thus, the molarity of the solution is 1.26 × 10^-8 M

6 0

3 years ago

When Mendeleev started his work on organizing the elements, about how many had

Talja [164]

Answer:

About 60.

Explanation:

Mendeleev knew of 63 elements. He wrote their properties on cards and arranged them in order of atomic mass.

That's how he discovered that the properties of elements are periodic functions of their atomic masses.

8 0

3 years ago

Acidic solutions contain high concentrations of

finlep [7]

Answer:

hydrogen ions

Explanation:

because acid is the specie that have ability to donate proton or forming bond with electron pair

6 0

3 years ago

The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall

miv72 [106K]

Answer:

The value is $C_p = 42. 8 J/K\cdot mol$

Explanation:

From the question we are told that

$\gamma = \frac{C_p }{C_v}$

The initial volume of the fluorocarbon gas is $V_1 = V$

The final volume of the fluorocarbon gas is $V_2 = 2V$

The initial temperature of the fluorocarbon gas is $T_1 = 298.15 K$

The final temperature of the fluorocarbon gas is $T_2 = 248.44 K$

The initial pressure is $P_1 = 202.94\ kPa$

The final pressure is $P_2 = 81.840\ kPa$

Generally the equation for adiabatically reversible expansion is mathematically represented as

$T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }$

Here R is the ideal gas constant with the value

$R = 8.314\ J/K \cdot mol$

$248.44 = 298.15 * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }$

=> $C_v = 31.54 J/K\cdot mol$

Generally adiabatic reversible expansion can also be mathematically expressed as

$P_2 V_2^{\gamma} = P_1 V_1^{\gamma}$

=> $[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}$

=> $2^{\gamma} = 2.56$

=> $\gamma = 1.356$

$\gamma = \frac{C_p}{C_v} \equiv 1.356 = \frac{C_p}{31.54}$

=> $C_p = 42. 8 J/K\cdot mol$

3 0

3 years ago

Please answer PLEASE ILL GIVE BRAINLIEST

netineya [11]

C ! ..........................

3 0

3 years ago

Read 2 more answers