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
because it can influence how frequently and sufficiently the particles collide depending on the space it has to do so, for example a large surface area would be have a slower rate of reaction and a lower temperature. (the rate of reaction in terms of concentration, it is diffused from high to low)
They all have densities greater than the density of the fluid in which they are<span>sinking. The mass of the displaced liquid is less than the mass of the sinking body.</span>
Answer:
"Protecting the privacy of your employees is critical to the long term success of your business. Senior leadership, high-profile executives and their families, as well as other customer-facing employees are particularly vulnerable to targeting to bad actors when their personal information is exposed. Other potential sources of danger include disgruntled employees, ex-employees and even customers."
Hope this helps :)
