Answer:
see explaination
Explanation:
Molecular equation;
2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)
Total ionic equation; . Includes all ions ;
6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)
Net ionic equation; remove common ions from total ionic;
2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)
Answer:
d = 0.793 g/L
Explanation:
Given data:
Density of fluorine gas = ?
Pressure of gas = 0.554 atm
Temperature of gas = 50 °C (50+273.15K = 323.15 K)
Solution:
Formula:
PM = dRT
M = molar mass of gas
P = pressure
R = general gas constant
T = temperature
d = PM/RT
d = 0.554 atm × 37.99 g/mol / 0.0821 atm.L /mol.K × 323.15 K
d = 21.05 atm.g/mol/26.53 atm.L /mol
d = 0.793 g/L
Answer:
- <u><em>No, I would not consider a metal to be a plasma because plasma is just another state of matter, and the copper wire is in solid state.</em></u>
Explanation:
Metal is not a state of matter. Metals can be solid or liquid (molten) depending on their melting point and the temperature at which they are.
Plasma is a state of matter, similar to gas, but it is reached only at very high temperatures like in the Sun. The particles in plasma state are not neutral atoms or molecules but negatively charged ions and electrons.
The copper wire is yet a solid, thus it cannot be considered a plasma.
Metals can be in plasma state only if the temperature is too high, like the temperatures in the stars. In fact, the metals in the Sun and other hotter stars are in plasma state.
<h3>
Answer:</h3>
0.0157 g Au
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.113 g Au
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 197.87 g/mol
<u>Step 3: Convert</u>
<u />
= 0.015733 g Au
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.015733 g Au ≈ 0.0157 g Au
