Answer:
Therefore the density of the sheet of iridium is 22.73 g/cm³.
Explanation:
Given, the dimension of the sheet is 3.12 cm by 5.21 cm.
Mass: The mass of an object can't change with respect to position.
The S.I unit of mass is Kg.
Weight of an object is product of mass of the object and the gravity of that place.
Density: The density of an object is the ratio of mass of the object and volume of the object.
[S.I unit of mass= Kg and S.I unit of m³]
Therefore the S.I unit of density = Kg/m³
Therefore the C.G.S unit of density=g/cm³
The area of the sheet is = length × breadth
=(3.12×5.21) cm²
=16.2552 cm²
Again given that the thickness of the sheet is 2.360 mm =0.2360 cm
Therefore the volume of the sheet is =(16.2552 cm²×0.2360 cm)
=3.8362272 cm³
Given that the mass of the sheet of iridium is 87.2 g.
=22.73 g/cm³
Therefore the density of the sheet of iridium is 22.73 g/cm³.
Answer:
H₃PO₄/H₂PO₄⁻ and HCO₃⁻/CO₃²⁻
Explanation:
An acid is a proton donor; a base is a proton acceptor.
Thus, H₃PO₄ is the acid, because it donates a proton to the carbonate ion.
CO₃²⁻ is the base, because it accepts a proton from the phosphoric acid.
The conjugate base is what's left after the acid has given up its proton.
The conjugate acid is what's formed when the base has accepted a proton.
H₃PO₄/H₂PO₄⁻ make one conjugate acid/base pair, and HCO₃⁻/CO₃²⁻ are the other conjugate acid/base pair.
H₃PO₄ + CO₃²⁻ ⇌ H₂PO₄⁻ + HCO₃⁻
acid base conj. conj.
base acid
If there are options, they are b, c, and d. I've done this question in class before :)
<span>In order to do this, you have change the alkene into an
alkyne. That is the aim of Br2/CH2Cl2 trailed by NaNH2. The Br2 with form a vic
dihalide (3,4-dibromo octane). Adding of NaNH2 will execute two E2 reactions.
-NH2 will eliminate an H from carbons 3 and 4. This double elimination will make
the alkyne. Then handling the alkyne with H2/Lindlar will form the cis alkene. The
final product will be CIS-3-octene.</span>
Let us differentiate accuracy from precision. Accuracy is the nearness of the measured value to the true or exact value. On the other hand, precision is the nearness of the measured values between each other. So, for precision, select the student in which the measured values are very near to each other. That would be Student III. Now, for accuracy, let's find the average for each student.
Student I: (<span>8.72g+8.74g+8.70g)/3 = 8.72 g
Student II: (</span><span>8.56g+8.77g+8.83g)/3 = 8.72 g
Student III: (</span><span>8.50g+8.48g+8.51g)/3 = 8.50 g
Student IV: (</span><span>8.41g+8.72g+8.55g)/3 = 8.56 g
From the given results, the accurate one would be Students I and II. So, we make a compromise. Even though Student III is precise, it is not accurate. If you compare between Students I and II, the more precise data would be Student I. Therefore, the answer is Student I.</span>
