Answer : The question is to write the elements alphabetically and seperating them with commas.
So in this case for your question of Fireworks , K
,
The answer will be
K, N, O.
You have to just differentiate the elements and write them according to the alphabetical order.
Answer:
A. P₂ / P₁ = 2
B. P₂ / P₁ = 1.1
Explanation:
A. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 46 K
Final temperature (T₂) = 92 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/46 = P₂/92
Cross multiply
46 × P₂ = P₁ × 92
Divide both side by P₁
46 × P₂ / P₁ = 92
Divide both side by 46
P₂ / P₁ = 92 / 46
P₂ / P₁ = 2
B. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K
Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/308.4 = P₂/342
Cross multiply
308.4 × P₂ = P₁ × 342
Divide both side by P₁
308.4 × P₂ / P₁ = 342
Divide both side by 308.4
P₂ / P₁ = 342 / 308.4
P₂ / P₁ = 1.1
Answer:
LR is Na₃PO₄
Explanation:
A quick way to determine the limiting reactant in a process is to convert reactant values to moles and then divide by the respective coefficient of the balanced equation. The smaller number of the division is the limiting reactant. For the given reaction, the rxn ratio of reactants is 1:1 so only the smaller mole value gives limiting reactant. However, if the reaction is NOT 1:1 the one must divide by respective coefficient to identify the smallest value and the limiting reactant.
This problem:
FeCl3(aq) + Na3PO4(aq) => FePO4(s) + 3 NaCl(aq)
Given: 27.8g 61.9g
moles: 27.8g/162.2g/mole 1.9g/163.94g/mole
= 0.1714 mole = 0.0116 mole
÷ coef. => 0.1714/1 = 0.1714 => 0.0116/1 = 0.0116
smaller value is LR => => => => => => LR is Na₃PO₄
Hope this helps. Doc :-)
Answer:
Enter your numerical answer to 1 decimal place. Do not include the units of "g". Question: Question 14 1 pts Consider the following reaction, where A and E are hypothetical elements. A+3E --> AE The atomic mass of A is 42.1 amu. The atomic mass of Eis 56.2 amu. If 42.6 g of A fully reacts, how many grams of AE, are expected to form?
The charge on Pb in Pb(SO3)2 is Lead (IV) Sulfite.