Ok cool I ujust do you think I should be able to get a hold with him tomorrow morning or t if rry want me too early or something I don’t know how to get it off or if you want me too I just need it for a while and I’ll get back with him
Knowing the ratio between atoms we can write an empirical formula:
<span>C4H6O </span>
<span>we compute the molar mass of this single formula: </span>
<span>4x12 + 6 x 1 + 16 x1 = 70 g / mol </span>
<span>Now, as we know the actual molar mas being 280 g/mol, we divide this number by 70 and we get the ratio between empirical formula and molecular actual formula: </span>
<span>280 / 70 = 4 </span>
<span>This means that actual molecular formula is: </span>
<span>(C4H6O)4 or </span>
<span>C16H24O4 </span>
Theyre temporary magnets, so they can be turned on and off.
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
