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irina [24]
2 years ago
10

5. What is the mass of 9.80 x 1023 formula units of zinc chlorate, Zn(CO3)2?

Chemistry
1 answer:
Zielflug [23.3K]2 years ago
5 0

Answer:

\boxed{\text{378 g}}

Explanation:

We must convert formula units of Zn(ClO₃)₂ to moles and then to grams of Zn(ClO₃)₂.

Step 1. Convert formula units to moles

\text{Moles of Zn(ClO$_{3}$)$_{2}$}\\\\= 9.80 \times10^{23}\text{ formula units Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}{6.022 \times\ 10^{23} \text{ formula units Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$}

Step 2. Convert moles to grams

\text{Mass of Zn(ClO$_{3}$)$_{2}$}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{232.29 g Zn(ClO$_{3}$)$_{2}$}}{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{378 g Zn(ClO$_{3}$)$_{2}$}\\\\\text{The mass of Zn(ClO$_{3}$)$_{2}$ is } \boxed{\textbf{378 g}}

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HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

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C_iVi = C_fV_f

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V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

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Explanation :

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As there are 14 electrons present in the given configuration.

The molecular orbital configuration of molecule will be,

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