Convert 57.6 L to dm3 and divide it by 24
Identify each of the following as a product or a coefficient in reaction below
<h2> 2H^+ + CO3^2- → H2O + CO2</h2><h2 /><h2 /><h3> <u><em>reactant are</em></u></h3><h2> H^+ and CO3^2-</h2>
<u><em>Reason: </em></u>Reactant of a chemical reaction are found in the left side. They are initially present in a chemical reaction which are consumed to form product.
<h3> <em><u>Product are</u></em></h3>
H2O and CO2
<u><em>Reason</em></u>: They are found in the right side of the reaction. Product are produced in a chemical reaction.
<em><u>coefficient</u></em>
<em><u> </u></em><em><u> </u></em>is 2
- Coefficient is the number found in front of a formula.
- Therefore 2 is the coefficient since it is found in front of H^+
Answer:
B. –99 kJ.
Explanation:
We have the following information:
1. C(s) + O₂(g) → CO₂(g);
ΔH = -393 kJ
2. 2CO(g) + O₂ → 2CO₂(g);
ΔH = -588 kJ
Using Hess's Law, Our target equation has C(s) on the left hand side, so we re-write equation 1:
1. C(s) + O₂(g) → CO₂(g);
ΔH = -393 kJ
So, we reverse equation 2 and divide by 2, we have equation 3:
3. CO₂(g) → CO(g) + ½O₂;
ΔH = +294 kJ
That is, change the sign of ΔH and divide by 2. Then we add equations 1 and 3 and their ΔH values.
This gives:
C(s) +½O₂(g) → CO(g);
ΔH = +294 - 393 kJ
= -99 kJ
The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.
Answer: The weight of the products should be the same after he mixes them.
Explanation: The law of the conservation of mass states that the mass of the products in a chemical reaction must equal the mass of the reactants. Even if the products are changed or rearranged, the products and reactants will still have the same mass
Answer:
64.0
Explanation:
2Mg+O2 ---> 2MgO
use dimentional analysis to find the amount of moles of O2 needed first
4.00molMg x 1.00mol O2/ 2.00 mol Mg=. 2.00 mol O2
using the coefficients you can see the mole ratio for O2:Mg the mole ratio is 1:2 which is why there is 1 mole on the top for 2 moles on the bottom. The Mg would cancel and multiply 4 by 1 then divide by 2, or multipy 4 by 1/2
Now that you have the moles of O2 you use the molar mass to find the grams in 2 moles of O2
2.00 mol O2 x 32.0g/1.00 mol = 64.0 g
multiply 2 by 32
