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A sample of MgCO3 contains 3.01 xx 10^(2...

A sample of `MgCO_3` contains `3.01 xx 10^(23) Mg^(2+)` ions and `3.01 xx 10^(23) CO_(3)^(2+)` ions. The mass of the sample is:

A

42 mg

B

84 g

C

0.042 kg

D

42 mol

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To solve the problem of finding the mass of the sample of \( \text{MgCO}_3 \) containing \( 3.01 \times 10^{23} \) \( \text{Mg}^{2+} \) ions and \( 3.01 \times 10^{23} \) \( \text{CO}_3^{2-} \) ions, we can follow these steps: ### Step 1: Determine the number of moles of \( \text{Mg}^{2+} \) ions We know that 1 mole of any substance contains \( 6.022 \times 10^{23} \) particles (Avogadro's number). \[ \text{Number of moles of } \text{Mg}^{2+} = \frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.5 \text{ moles} \] ### Step 2: Calculate the mass of \( \text{Mg}^{2+} \) The molar mass of \( \text{Mg}^{2+} \) is approximately 24 g/mol. \[ \text{Mass of } \text{Mg}^{2+} = \text{Number of moles} \times \text{Molar mass} = 0.5 \times 24 = 12 \text{ grams} \] ### Step 3: Determine the number of moles of \( \text{CO}_3^{2-} \) ions Since the number of \( \text{CO}_3^{2-} \) ions is the same as the \( \text{Mg}^{2+} \) ions, we also have: \[ \text{Number of moles of } \text{CO}_3^{2-} = 0.5 \text{ moles} \] ### Step 4: Calculate the mass of \( \text{CO}_3^{2-} \) The molar mass of \( \text{CO}_3^{2-} \) can be calculated as follows: - Carbon (C): 12 g/mol - Oxygen (O): 16 g/mol (and there are 3 oxygen atoms) \[ \text{Molar mass of } \text{CO}_3^{2-} = 12 + (3 \times 16) = 12 + 48 = 60 \text{ g/mol} \] Now, calculate the mass of \( \text{CO}_3^{2-} \): \[ \text{Mass of } \text{CO}_3^{2-} = \text{Number of moles} \times \text{Molar mass} = 0.5 \times 60 = 30 \text{ grams} \] ### Step 5: Calculate the total mass of the sample Now, we can find the total mass of the sample by adding the masses of \( \text{Mg}^{2+} \) and \( \text{CO}_3^{2-} \): \[ \text{Total mass} = \text{Mass of } \text{Mg}^{2+} + \text{Mass of } \text{CO}_3^{2-} = 12 + 30 = 42 \text{ grams} \] ### Step 6: Convert grams to kilograms (if needed) To convert grams to kilograms, we divide by 1000: \[ \text{Total mass in kg} = \frac{42}{1000} = 0.042 \text{ kg} \] ### Final Answer The mass of the sample is **42 grams** or **0.042 kilograms**. ---

To solve the problem of finding the mass of the sample of \( \text{MgCO}_3 \) containing \( 3.01 \times 10^{23} \) \( \text{Mg}^{2+} \) ions and \( 3.01 \times 10^{23} \) \( \text{CO}_3^{2-} \) ions, we can follow these steps: ### Step 1: Determine the number of moles of \( \text{Mg}^{2+} \) ions We know that 1 mole of any substance contains \( 6.022 \times 10^{23} \) particles (Avogadro's number). \[ \text{Number of moles of } \text{Mg}^{2+} = \frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.5 \text{ moles} \] ...
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