The volume of the textbook calculated by the student written to 3 significant figures is 992cm³
- According to this question, a student calculates the volume of his textbook and arrived at an answer of 991.880 cm³.
- To round up this digit to 3 significant figures, which means that only 3 significant numbers should appear in the answer.
- To do this, the number after the third number is 8, which means we can round up the value of 1 to 2. Hence, the volume of the textbook calculated by the student is 992cm³ in 3 s.f.
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Answer:
7mL of sterile water is the initial amount of the concentrated solution is 3mL
Explanation:
In this problem, the vial must be <em>diluted </em>from 5mg/mL to 1.5mg/mL, that means the solution must be diluted:
5mg/mL / 1.5mg/mL = 3.33 times
If the initial amount of the drug in the vial is 3mL, the final volume must be:
10mL
That means the volume of water that should be added is:
10mL - 3mL:
<h3>7mL of sterile water is the initial amount of the concentrated solution is 3mL</h3>
In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.
Explanation:
In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.
I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.
It can be said that only 38.3% of energy is put in ATP molecules.
Answer:
10043.225 J
Explanation:
We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:
Mass (m) = 15.5g
Latent heat of fussion of water (L) = 334J/g
Heat (Q1) =..?
Q1 = mL
Q1 = 15.5 x 334
Q1 = 5177 J
Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.
This is illustrated below:
Mass = 15.5g
Initial temperature (T1) = 0°C
Final temperature (T2) = 75°C
Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C
Specific heat capacity (C) of water = 4.186J/g°C
Heat (Q2) =?
Q2 = MCΔT
Q2 = 15.5 x 4.186 x 75
Q2 = 4866.225 J
The overall heat energy needed is given by:
QT = Q1 + Q2
QT = 5177 + 4866.225
QT = 10043.225 J
Therefore, the amount of energy required is 10043.225 J
Answer:
Explanation:
Hello there!
In this case, according to the described chemical reaction, it is possible to compute the theoretical mass of silver as mass via the 1:2 mole ratio of copper to silver and their atomic mass in the periodic table, in order to perform the following stoichiometric setup:
Next, given the actual yield of 120 g, we compute the percent yield via:
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