Answer:
Warm front Forms when a moist, warm air mass slides up and over a cold air mass. As the warm air mass rises, it condenses into a broad area of clouds. A warm front brings gentle rain or light snow, followed by warmer, milder weather.
Your answer is B.
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Answer:
28 g/mol, N2
Explanation:
Given data:
Volume of gas = 5.0 L
Mass of gas = 6.3 g
Pressure = 1 atm
Temperature = 273 K
Molar mass of gas = ?
Solution:
We will calculate the density first.
d = mass/ volume
d = 6.3 g/ 5.0 L
d = 1.26 g/L
Molar mass:
d = PM/RT
M = dRT/P
M = 1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm
M = 28 g/mol
Molar mass of N₂ is 28 g/mol thus given gas is N₂.
Answer: Please see answer below
Explanation:
Mecury vapor lamp is better to use than Sodium vapor light, this is because because
---The Filaments of the lamp in sodium emit fast moving electrons, which causes valence electrons of the sodium atoms to excite to higher energy levels which when electrons after being excited, relax by emitting yellow light which concentrates on the the monochromatic bright yellow part of the visible spectrum which is about 580-590 or about (589nm) which will fall incident on the calibrations making it difficult to see
While
In Mercury vapor lamp, The emitted electrons from the filaments, after having been excited by high voltage, hit the mercury atoms but the excited electrons of mercury atoms relax and emits an ultraviolet uv invisible lights falling on the mecury vapour lamp to produce white light covering a wide range of (380-780 nm) which is visible that is why it is used for calibrations purposes in lightening applications.
The balanced equation is Mg + 2AgNO₃ ⟶ Mg(NO₃)₂ + 2Ag
Step 1. Write the <em>unbalanced equation
</em>
Mg + AgNO₃ ⟶ Mg(NO₃)₂ + Ag
Step 2. Start with the<em> most complicated-looking formula</em> [Mg(NO₃)₂] and balance its atoms.
Mg: Already balanced —1 atom each side.
N: We need 2 N on the left. Put a 2 in front of AgNO₃.
1Mg + 2AgNO₃ ⟶ 1Mg(NO₃)₂ + Ag
O: Already balanced —6 atom6 each side.
Step 3: Balance <em>Ag</em>
We have 2Ag on the left. We need 2Ag on the right.
1Mg + 2AgNO₃ ⟶ 1Mg(NO₃)₂ + 2Ag