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2 years ago

Calculates the equivalent mass of calcium chlorate CA (CI03) 2

Chemistry

1 answer:

love history [14]2 years ago

3 0

Answer:

207g

Explanation:

Given compound:

Ca(ClO₃)₂

The equivalent mass can be derived by summing the molar masses of each atom

Molar mass of Ca = 40

Cl = 35.5

O = 16

Now solve;

Molar mass = 40 + 2(35.5 + 3(16)) = 207g

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How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?

raketka [301]

<h2>Answer:</h2>

n = 0.0989 moles

<h2>Explanation:</h2>

n = PV / RT

P = 2.09atm

V = 1.13L

R = 0.08206

T = 291K

Plug the numbers in the equation.

n = (2.09atm)(1.13L) / (0.08206)(291K)

n = 0.0989 moles

3 0

3 years ago

If the value for ΔS is postive, and the value for ΔH is negative, thr reaction will be

dmitriy555 [2]

It follows that the reaction is spontaneous at high temperatures Option A.

The term ΔS is referred to as the change in the entropy of the system. Now recall that entropy is defined as the degree of disorderliness in a system. If a system is highly disorderly then it means that it has a high entropy. Also, ΔH has to do with the heat change that accompanies a reaction.

We know that both the entropy and the heat change can both either be positive or negative. Now we know that the equation ΔG = ΔH - TΔS can be used to ascertain whether or not a reaction will be spontaneous. If the result is negative, then the reaction will be spontaneous.

As such, when then it follows that the reaction is spontaneous at high temperatures Option A.

Learn more about spontaneous reaction:brainly.com/question/13790391

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6 0

1 year ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$