Answer:
342.8 kJ are absorbed
Explanation:
In the reaction:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = 1168 kJ
<em>As ΔH > 0, the heat is absorbed. Also, when 4 moles of NH3 are involved in the reaction, there are absorbed 1168 kJ</em>.
Having this in mind, moles of NH3 in 20.00g are:
20.00g × (1mol / 17.0307g) = <em>1.174 moles</em>
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Thus, 1.174 moles of NH3 absorbed:
1.174 moles × (1168 kJ / 4 moles) = <em>342.8 kJ are absorbed</em>.
Answer:
mass of sodium reacted is 184.1 g
Explanation:
mass Na = X = ?
∴ mass NaCl = 468 g
∴ mass Cl = 0.248 g
∴ molar mass NaCl = 58.44 g/mol
∴ atomic mass Cl = 35.453 a.m.u
∴ atomic mass Na = 22.989 a.m.u
⇒ moles Na = (X gNa)*(mol Na/22.989 g) = X/22.989 mol Na
⇒ mass NaCl = (X/22.989 mol Na)*(mol NaCl/mol Na)*(58.44 gNaCl/mol NaCl) = 468 g NaCl
clearing "X":
⇒ ((58.44)(X))/(22.989) = 468 g
⇒ X = 184.1 g = mass Na reacted
Answer:
5.67 g OF WATER WILL BE FORMED WHEN 13.7 g OF MnO2 REACTS WITH HCl GAS.
Explanation:
EQUATION FOR THE REACTION
Mn02 + 4HCl --------> MnCl2 + Cl2 + 2H2O
From the balanced reaction between manganese oxide and hydrogen chloride gas;
1 mole of MnO2 reacts to form 2 mole of water
At STP, the molecular mass of the sample is equal to the mole of the substance. So therefore:
(55 + 16 * 2) g of MnO2 reacts to form 2 * ( 1 *2 + 16) g of water
(55 + 32) g of MnO2 reacts to form 2 * 18 g of water
87 g of MnO2 reacts to form 36 g of water
If 13.7 g of MnO2 were to be used?
87 g of MnO2 = 36 g of H2O
13.7 g of MnO2 = ( 13.7 * 36 / 87) g of water
= 493.2 / 87 g of water
Mass of water = 5.669 g of water
Approximately 5.67 g of water will be formed when 13.7 g of manganese oxide reacts with excess hydrogen chloride gas.
When perfume or cologne is sprayed into the air, it turns into a gas and its particles mix with other air particles that quickly move around the room. This is because gas is the least dense of the three states of matter. It moves quickly into any space or volume because it has the least density. This is called diffusion, when molecules or particles move from areas with high concentration to low concentration areas.
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