Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
Answer:
The temperature change from the combustion of the glucose is 6.097°C.
Explanation:
Benzoic acid;
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.570 g
Moles of benzoic acid = 
Energy released by 0.004667 moles of benzoic acid on combustion:
Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.053°C
Glucose:
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=2.900 g
Moles of glucose = 
Energy released by the 0.016097 moles of calorimeter combustion:
Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'
The temperature change from the combustion of the glucose is 6.097°C.
Answer:
- <em>Oxidation half-reaction</em>:
Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
- <em>Reduction half-reaction</em>:
Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
Explanation:
The reaction that takes place is:
- Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)
The <em>oxidation half-reaction</em> is:
- Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.
The <em>reduction half-reaction</em> is:
- Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.
Answer: This is from a wiki i found. Approximately one third of a cell’s proteins are destined to function outside the cell’s boundaries or while embedded within cellular membranes. Ensuring these proteins reach their diverse final destinations with temporal and spatial accuracy is essential for cellular physiology. In eukaryotes, a set of interconnected organelles form the secretory pathway, which encompasses the terrain that these proteins must navigate on their journey from their site of synthesis on the ribosome to their final destinations. Traffic of proteins within the secretory pathway is directed by cargo-bearing vesicles that transport proteins from one compartment to another. Key steps in vesicle-mediated trafficking include recruitment of specific cargo proteins, which must collect locally where a vesicle forms, and release of an appropriate cargo-containing vessel from the donor organelle (Figure 1). The newly formed vesicle can passively diffuse across the cytoplasm, or can catch a ride on the cytoskeleton to travel directionally. Once the vesicle arrives at its precise destination, the membrane of the carrier merges with the destination membrane to deliver its cargo. Have a nice day.
Explanation: Plz make brainliest
