Answer:
0.26g of NaCl is the maximum mass that could be produced
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H₂O
<em>Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl</em>
<em />
To solve this question we need to find <em>limiting reactant. </em>The moles of limiting reactant = Moles of NaCl produced:
<em>Moles HCl -Molar mass: 36.46g/mol-:</em>
0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl
<em>Moles NaOH -Molar mass: 40g/mol-:</em>
0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH
As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.
The mass of NaCl is:
<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>
0.0045 moles * (58.44g/mol) =
<h3>0.26g of NaCl is the maximum mass that could be produced</h3>
Answer:
The correct answer is B. exogenous
Explanation:
Let us try to describe exogenous and endogenous variables an exogenous variable value is influenced only by factors outside a model or system and is forced onto the model, while a change in an exogenous variable is known as an exogenous change. Also an endogenous variable is one whose value is influenced only by the system or model under study.
Answer:
1. V2.
2. 299K.
3. 451K
4. 0.25 x 451 = V2 x 299
Explanation:
1. The data obtained from the question include:
Initial volume (V1) = 0.25mL
Initial temperature (T1) = 26°C
Final temperature (T2) = 178°C
Final volume (V2) =.?
2. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K
3. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Final temperature (T2) = 178°C
Final temperature (T1) = 178°C + 273 = 451K
4. Initial volume (V1) = 0.25mL
Initial temperature (T1) = 299K
Final temperature (T2) = 451K
Final volume (V2) =.?
V1 x T2 = V2 x T1
0.25 x 451 = V2 x 299
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
