The density of a substance is simply defined as the mass of the subtance per unit volume of the substance.
To obtain the density of the aluminum, we'll begin by calculating the volume of the aluminum. This can be obtained as follow:
From the question given above, each side of the aluminum cube measures 3 cm. Thus,
Length (L) = 3 cm
Width (W) = 3 cm
Height (H) = 3 cm
Volume = Length × Width × Height
Volume = 3 × 3 × 3
Therefore, the volume of the aluminum is 27 cm³
Finally, we shall determine the density of the aluminum. This can be obtained as follow:
Mass = 81 g
Volume = 27 cm³
Therefore, the density of the aluminum is 3 g/cm³
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Answer:
27
Explanation: volume-3 times 6=18 /mass- 81 times 6= 486.... 486 divided by 18 = 27.... a cube has 6 sides.
The balanced chemical equation is :
Moles of ,
Now, by given chemical equation , we can see 2 mole of react with 1 mole of .
So , 2.27 mole react with :
Mass of oxygen is :
Therefore, mass of oxygen in grams produced is 18.16 g.
Hence, this is the required solution.
Moles of any substance is given by the molar mass and the mass. The volume of EDTA required to titrate is 3.16 mL and the volume of zinc required to titrate is 22.8 mL.
Molarity is the property of the solution that gives the concentration of the solute present in the solution.
Given,
Mass of cobalt sulfate = 1.697 gm
The molar mass of cobalt sulfate = 155 g/mol
In the first part, the volume of the aliquot is 25 mL and the molarity is 0.08640 M.
The reaction is shown as:
Co²⁺ + H₄Y → CoH₂Y + 2H⁺
Moles of cobalt: n = 0.001694 ÷ 155 = 0.0000109 moles
In 25 ml aliquot moles of cobalt are, 0.000273 moles.
The volume of EDTA solution is calculated as:
V = moles ÷ Molarity
= 0.000273 mole ÷ 0.0864 mol/ L
= 3.16 mL
Hence, 3.16 mL of EDTA is required.
For the second part, moles of EDTA are calculated as:
n = 0.008640 × 0.050 = 4.32 ×10⁻⁴
In a 25 mL, sample moles of EDTA are 2.16 × 10⁻⁴ moles.
Excess moles of EDTA in the solution:
0.000432 - 0.000216 = 0.000216 moles
The volume of EDTA is calculated as:
V = 0.000216 ÷ 0.009450
= 0.0228 L
Hence, 22.8 mL of zinc is required.
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Answer:
the answer is in the screenshot
Explanation:
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The answer I got was False, is this correct?
Answer: yes it is false
Explanation:
The statement is false. A synthetic process with a lower E-factor produces less waste than a process with a higher E-factor.
The E-factor is a measure of the waste generated during a manufacturing process. It is calculated by dividing the total mass of waste produced by the mass of the desired product. A lower E-factor indicates that less waste is generated per unit of product.
In this case, the synthetic process with an E-factor of 3.0 produces less waste than the process with an E-factor of 17.4. This means that the process with an E-factor of 3.0 is more efficient in terms of waste reduction.
B) an alpha particle or a helium atom.
C) a beta particle or a hydrogen nucleus.
D) an alpha particle or a helium nucleus.
The radioactive uranium decays to produce thorium and it emits an alpha particle or helium atom. Thus, option A is correct.
Unstable heavy isotopes of elements undergo nuclear decay to produce stable atoms by the emission of charged particle such as alpha or beta particles.
Based on the emitted particle, there are two types of decay process namely alpha decay and beta decay. In alpha decay atoms emits alpha particles which are helium nuclei and the atom losses its mass number by 4 units and atomic number by two units,
In beta decay, electrons are emitted by the atom, where no change occurs in mass number and atomic number increases by one unit. Uranium undergo alpha decay by emitting alpha particle or helium nuclei.
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Answer:
Option 3.
Explanation:
Isomerism is a phenomenon where by two or more compounds have the same molecular formula but different structural patterns.
Geometric Isomerism is a type of Isomerism that occurs within a double bond i.e Geometric isomers have different arrangement within the double bond.
Considering the options given above,
The 1st option is exactly the same as the compound only, it is inverted.
The 2nd option is still the same as the compound, only it is laterally inverted.
The 3rd option satisfy geometric Isomerism as the arrangement differ from the compound in the double bond.
The 4th option is entirely a saturated compound in which geometric Isomerism is not possible.