C. "A" mentions that the drug killed all of the bacteria that are resistant to the drug, and that doesn't make any sense. "B" claims that all bacteria are resistant to the drug. This is not true. "D" mentions that bacteria eat the drug, which doesn't happen.
Answer:
This question is incomplete. A fragment of the question is missing. The fragment is:
She finds that 4.87 g of oxygen gas is produced
0.152 moles
Explanation:
Using the formula: mole = mass / molar mass
According to the question, a chemist measures the amount of oxygen gas produced during an experiment. Oxygen gas has the chemical formula: O2. Hence, the molar mass will be:
O2 = 16(2) = 32g/mol
If the mass of the oxygen gas to be 4.87 g of oxygen gas.
Therefore,
moles = 4.87/32
moles = 0.152mol
Answer:
The enthalpy change for the solution process
= - 158.34 kJ/mol
Explanation:
Given that:
The mass of salt AX = 2.499 g
The mass of water = 135.3 g
The mass of the solution = ( 2.499 + 135.3 ) g = 137.799 g
The specific heat of salt solution s is known to be = 4.18 J/g° C
The change in temperature i.e. ΔT = 35.2 °C - 23.6 °C = 11.6 °C
Thus, the amount of heat raised is equal to the heat absorbed by the calorimeter.
∴





Recall that the mass of the salt = 2.499 g
The number of moles of the salt = 
= 0.0422 mol of AX
Finally the enthalpy change, 
= - 158.34 kJ/mol
The enthalpy change for the solution process
= - 158.34 kJ/mol
The standard heat of formation is the heat released or absorbed when you obtain the CrO₃ compound. So, the reaction basically involves the individual elements of the compound. The complete balanced chemical reaction for the formation of CrO₃ is:
<em>2 Cr + 3 O₂ → 2 CrO₃</em>
<h3><u>Answer</u>;</h3>
C3H4O
<h3><u>Explanation;</u></h3>
Empirical formula is the simplest formula of a compound;
Molar mass CO2 = 44.01
Mass of CO2 produced = 2.053 g
Mass of carbon in original sample = 12.01/44.01 × 2.053
= 0.5603g
Molar mass H2O = 18
Mass of H in original sample = 2/18 ×0.5601
= 0.0622 g
Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O
Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g
The mole ratio of C:H:O will be;
Moles C = 0.5603/12 = 0.0467
Moles H = 0.0622
Moles O = 0.249/16 = 0.01556
C:H:O = 0.0467:0.0622:0.01556
Divide through by 0.01556:
C:H:O = 3:4:1
Empirical formula is thus C3H4O