Answer:
q = 14049 J
Explanation:
q = m*c*(t2-t1)
q = 350 * 0.892 * (70-25) =
312.2 * 45 = 14049 J
I might be getting a little confused but I could be right.
Hope this helps!
Answer:
It's A
Explanation:
Liquid can heat things and so can steam.
Lets assume x volume of NaOH and x volume of HCl are added together.
NaOH ---> Na⁺ + OH⁻
NaOH is a strong base therefore it completely ionizes and releases OH⁻ ions into the medium
HCl ---> H⁺ + Cl⁻
HCl is a strong base and completely ionizes and releases H⁺ ions in to the medium. number of NaOH moles in 1 L - 0.1 mol
Therefore in x L - 0.1 /1 * x = 0.1x moles of NaOH present
Similarly in HCl x L contains - 0.1x moles of HCl
H⁺ + OH⁻ ---> H₂O
Due to complete ionisation, 0.1x moles of H⁺ ions and 0.1x moles of OH⁻ ions react to form 0.1x moles of H₂O. Therefore all H⁺ and OH⁻are completely used up and yield water molecules.
Then at this point the H⁺ and OH⁻ ions in the medium come from the weak dissociation of water. This is equivalent to 1 x 10⁻⁷M
pH = -log [H⁺]
pH = -log [10⁻⁷]
pH = 7
pH is therefore equals to 7 which means the solution is neutral
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
Explanation:
The two half equations are;
3e + HNO3 → NO
S→ H2SO4 + 6e
When balancing half equations, we have to make sure the number of electrons gained is equal to the number of electrons lost.
<em>Which factor will you use for the top equation?</em>
We multiply by 2 to make the number of electrons = 6e
<em>Which factor will you use for the bottom equation?</em>
We multiply by 1 to make the number of electrons = 6e