Answer:
800.0 mL.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>
M before dilution = 12.0 M, V before dilution = 100.0 mL.
M after dilution = 1.5 M, V after dilution = ??? mL.
∵ (MV)before dilution of HCl = (MV)after dilution of HCl
∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)
<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>
Answer:
The suitable equation for this reaction is
2CO + O₂ -----> 2CO₂
Here, we are given that we have 2 grams of O₂
From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂
Moles of O₂:
2/32 = 1/16 moles
Therefore, the number of moles of CO₂ is twice the moles of O₂
Moles of CO₂ = 2 * 1/16
Moles of CO₂ formed = 1/8 moles
Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂
Mass of CO₂ formed = 44 * 1/8
Mass of CO₂ formed = 5.5 grams
Hence, option B is correct
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Answer:
Explanation:
Hello!
In this case, since the net ionic equations are ionic representations of the molecular equation in which the spectator ions (those at both reactants and products sides) are cancelled out, we first write the complete ionic equation for this reaction, considering that the solid silver chloride is not ionized due to its precipitation:
Whereas the nitrate and sodium ions are cancelled out for the aforementioned reason as they are the spectator ions, to obtain:
Which is the required net ionic equation.
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