<span>3Ca + 2AlPO4 -> 2Al + Ca3(PO4)2 . . .the coefficient for Ca is 3.</span>
Answer:
The chemical formula is I2F9
Explanation:
D. is the answer
hope i could help
Answer:
1.55 × 10²⁴ atoms Xe
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 57.5 L Xe at STP
[Solve] atoms Xe
<u>Step 2: Identify Conversions</u>
[STP] 22.4 L = 1 mol
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe
Answer:
The answer to your question is: 1538095.2 kg of NH3
Explanation:
MW HNO3 = 63 kg
MW NO2 = 46 kg
3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)
3(46) kg-------------- 2(63) kg
x --------------- 7600000 kg
x = 7600000 x 138/126 = 8323809.5 kg og NO2
MW NO = 30
2 NO(g) + O2(g)---2 NO2(g)
2(30) ------------------2(46)
x ---------------- 8323809.5 kg
x = 8323809.5 x 60/92 = 5428571.4 kg of NO
MW NH3 = 17 kg
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
4(17) -------------------- 4(30)
x ----------------------- 5428571.4
x = 5428571.4 x 34 / 120
x = 1538095.2 kg of NH3