Answer:
name: zinc
symbol: Zn
atomic number: 30
number of protons: 30
atomic weight: 65.38 atomic mass units
I hope this helps!
Answer:
pHe = 3.2 × 10⁻³ atm
pNe = 2.5 × 10⁻³ atm
P = 5.7 × 10⁻³ atm
Explanation:
Given data
Volume = 1.00 L
Temperature = 25°C + 273 = 298 K
mHe = 0.52 mg = 0.52 × 10⁻³ g
mNe = 2.05 mg = 2.05 × 10⁻³ g
The molar mass of He is 4.00 g/mol. The moles of He are:
0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol
We can find the partial pressure of He using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 3.2 × 10⁻³ atm
The molar mass of Ne is 20.18 g/mol. The moles of Ne are:
2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol
We can find the partial pressure of Ne using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 2.5 × 10⁻³ atm
The total pressure is the sum of the partial pressures.
P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm
The answer is A because the time wouldn't affect the weather that much depending on the circumstances.
Answer:
319.8 m/min
Explanation:
533 cm/s
We can convert 533 cm/s to m/min by doing the following:
First, we shall convert 533 cm/s to m/s. This can be obtained as illustrated below:
Recall:
100 cm/s = 1 m/s
Therefore,
533 cm/s = 533 cm/s /100 cm/s × 1 m/s
533 cm/s = 5.33 m/s
Finally, we shall convert 5.33 m/s to m/min. This can be obtained as follow:
1 m/s = 60 m/min
Therefore,
5.33 m/s = 5.33 m/s / 1 m/s × 60 m/min
5.33 m/s = 319.8 m/min
Therefore, 533 cm/s is equivalent to 319.8 m/min
Answer:
El termopar B presenta un mayor grado de dispersión y también es más preciso. ... (c) La estimación para T = 175 ° C es probablemente la más cercana al valor real, porque el ... (cm3). Flujo de masa. Velocidad. (kg / min). Diferencia. Duplicar. (Di). Yo y yo. 2. 1 ... atm de gas. 2. 2. 2 f. 3. 2 f f. 30 14,7 lb 20 pulg. 4 14,7 lb 24 pulg 392 lb 7,00 10 lb pulg.
