<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
Answer is: d. 3,2,3.
Balanced chemical reaction: CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O.
CH₃CH₂OH is ethanol.
O₂ is molecule of oxygen.
CO₂ is carbon(IV) oxide.
H₂O is water.
There are same number of atoms (oxygen, carbon and hydrogen) on both side of balanced chemical reaction:
2 atoms of carbon.
6 atoms of hydrogen.
7 atoms of oxygen.
Answer:
The area of the given rectangular index card = <u>9677.4 mm²</u>
Explanation:
Area is defined as the space occupied by a two dimensional shape or object. The SI unit of area is square metre (m²).
<u>The area of a rectangle</u> (A) = length (l) × width (w)
Given dimensions of the rectangle: Length (l) = 5.0 inch, Width (w) = 3.0 inch
Since, 1 inch = 25.4 millimetres (mm)
Therefore, l = 5 × 25.4 = 127 mm, and w = 3 × 25.4 = 76.2 mm
Therefore, <u>the area of the given rectangular index card</u> = A= l × w = 127 mm × 76.2 mm = <u>9677.4 mm²</u>
Answer:
The value of the carbon bond angles are 109.5 °
Explanation:
CH3CH2CH2OH = propanol . This is an alcohol.
All bonds here are single bonds.
Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will create four hybrid orbitals with similar characteristics.
Sp3- types have angles of 109.5 ° between the carbon - atoms.
This means that the value of the carbon bond angles are 109.5 °
Answer:
The effects of supercritical CO2 (SC-CO2) on the microbiological, sensory (taste, odour, and colour), nutritional (vitamin C content), and physical (cloud, total acidity, pH, and °Brix) qualities of orange juice were studied. The CO2 treatment was performed in a 1 litre capacity double-walled reactor equipped with a magnetic stirring system. Freshly extracted orange juice was treated with supercritical CO2, pasteurised at 90°C, or left untreated. There were no significant differences in the sensory attributes and physical qualities between the CO2 treated juice and freshly extracted juice. The CO2 treated juice retained 88% of its vitamin C, while the pasteurised juice was notably different from the fresh juice and preserved only 57% of its vitamin C content. After 8 weeks of storage at 4°C, there was no microbial growth in the CO2 treated juice.
