Answer:
Lower explosive limit (LEL) of ethanol = 3.3%
Explanation:
In the case of alcohol, ethanol presents certain fire hazards. Its momentary flash point is 55ºF (12.9ºC), while the momentary flash point of gasoline is -45ºF (-42.8ºC), and the E85 mixture ranges between -20ºF and -4ºF (between -28 , 9ºC and -20ºC), and has a wider range of flammability limits than gasoline. For emergency response teams, this implies that during a release of the typical ethanol / gasoline mixture, the fuel can be expected to behave like gasoline: It is heavier than air - as we mentioned earlier - and can produce vapors and form flammable mixtures in the air, under most environmental conditions.
General properties and comparison with other inflambles products:
Flash point momentary Gasoline = -45 ° F
<u>Ethanol</u> = 55 ° F
E 85 = between -20º and -4º F
<u>Flammability limits
</u>
Lower explosive limit (LEL) of ethanol = 3.3%
Upper Explosive Limit (UEL) = 19%
Lower explosive limit (LEL) of the mixture E 85 = 1.4%
Upper Explosive Limit (UEL) 85 = 19%
Lower explosive limit (LEL) of gasoline = 1.4%
Upper Explosive Limit (UEL) = 7.6%
They have a wider range than gasoline
Lesser. atomic number means proton number
Answer:
CO32−
Explanation:
We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.
When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;
2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)
Answer: (Chlorine-35)
Element X is Ca
Explanation:
Number of proton plus the number of electron gives the mass number.
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
