Answer:
The amount in grams of hydrogen gas produced is 0.551 grams
Explanation:
The parameters given are;
Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms
Chemical equation for the reaction is given as follows;
2K + 2H₂O 
KOH + H₂
Avogadro's number, 
, regarding the number of molecules or atom per mole is given s follows;
= 6.02 × 10²³ atoms/mole
Therefore;
The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles
2 moles of potassium produces one mole of hydrogen gas, therefore;
1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas
The molar mass of hydrogen gas = 2.016 grams
Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.
The amount in grams of hydrogen gas produced = 0.551 grams.
They are odourless, colourless monatomic gases with full outer shells
Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
I think it’s A I’m not 100% sure but I mean it’s worth a try
