Use PV =nRT
so P = nRT/V
= 1 mole(0.08205 L atm/K mol)(1000K) / 2 L
= 41 atm
Answer:
Iron
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Answer:
510 g NO₂
General Formulas and Concepts:
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
- Reading the Periodic Table
- Writing Compounds
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol
<u>Step 3: Use Dimensional Analysis</u>
<u />
= 511.901 g NO₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
511.901 g NO₂ ≈ 510 g NO₂
Answer:
C - no antibodies
Explanation:
I dont think there is any blood type without antibodies
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
