Gases
liquids
plasma
solids
condensates <span />
Answer:
Explanation:
A t-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups, which may be related in certain features. The t-test is one of many tests used for the purpose of hypothesis testing in statistics. Calculating a t-test requires three key data values.
Answer:
7.04 g
Explanation:
Let's consider the reaction in the last step of the Ostwald process.
3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)
The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 6.40 g are:
6.40 g × (1 mol/63.01 g) = 0.102 mol
The molar ratio of NO₂ to HNO₃ is 3:2. The reacting moles of NO₂ are:
0.102 mol HNO₃ × (3 mol NO₂/2 mol HNO₃) = 0.153 mol NO₂
The molar mass of NO₂ is 46.01 g/mol. The mass corresponding to 0.153 moles is:
0.153 mol × (46.01 g/mol) = 7.04 g
Answer: One carrier interested in taxonomy is a veterinarian
Explanation:
Veterinarians as said, are interested in taxonomy. They are interested with animal organisms of all types with all species of animals. Hope this helps. Let me know if you need anything else.
Answer:
333.7 g.
Explanation:
- The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>
Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).
Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>
molaity (m) is the no. of moles of solute per kg of the solvent.
∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))
∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.
∵ no. of moles = mass/molar mass.
<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>
