To find for the oxidizing agent, first let us write the
half reactions of this complete chemical reaction:
Ca = Ca2+ + 2e- <span>
2 H+ + 2e- = H2</span>
The oxidizing agent
would be the substance of the element that is reduced. We know that an element
is reduced when an electron is added to it. In this case, the element being
reduced is H. Therefore the oxidizing agent is HNO3.
Answer:
<span>HNO3</span>
Answer:
The Density Calculator uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). The calculator can use any two of the values to calculate the third. Density is defined as mass per unit volume.
Answer: -227 kJ
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%20n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BC_2H_2%7D%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%282%5Ctimes%20-393.5%29%2B%281%5Ctimes%20-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28%5Cfrac%7B5%7D%7B2%7D%5Ctimes%200%29%5D)
![-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%28-787%29%2B%28-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%280%29%5D)

Therefore, the enthalpy change for
is -227 kJ.
Answer:
Explanation:
The law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction. This can be gleaned from the third postulate in Dalton's series. Magnesium oxide decomposes into magnesium and oxygen. If 4.03 g of magnesium oxide decomposes to form 2.43 g of magnesium, what mass of oxygen gas is also released in the reaction
The word say magnesium oxide decomposes to magnesium and oxygen
the chemical symbols say
MgO-----------> Mg + O2 (since natural oxygen is diatomic)
the balanced equation says
2MgO-------------->2Mg + O2
4.03 gm----------> 2.43 + ?0 gms
tour high school Algebra I class says
? = 4.03 -2.43 =1.60
your chemical analytcal lab says %mO in MgO = 16/40.3 = 39.7%
your calculator says
39.7/100 X 4.03 = 1.60
all of these prove the law of conservation of mass