To answer the question above, let us a basis of the 1000 mL or 1 L.
volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%.
mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol).
n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
molarity = 1.08 mol/ 1 L = 1.08 M
hydrocarbon is ethene which is used to test for saturation and it undergoes addition reaction
Answer:
Natural resources are naturally formed of course(wood copper fruits) and they are not distributed evenly due to competition of getting them and the prices of the resources
Explanation:
Nanochemicals can be defined as chemicals generated by using nanomaterials (materials that possess of size on nanometer dimensions). The nanochemicals are used in multiple different applications including chemical warfare, bicycle making, armor design and military weapons crafting. The most commonly used and observed nanochemicals are carbon nanotubes that are used a ton in industry for applications such as stronger materials (stronger bicycles).
Smart materials are exquisitely designed materials whose property(ies) can be modified with the use of an external stimulus such as temperature, stress, pH, and so on. Some examples of smart materials include shape memory materials, piezoelectric materials, ferrofluids, self-healing materials, and such. Applications involve memory pillows, memory based solar panels (for satellites), light sensitive glasses, and so on.
Specialized materials are made specifically to perform a specified task or function. Applications involve electronic equipment (high purity silicon & germanium), machine tools (high tungsten high carbon steel), dental filling (dental amalgam), and so on.
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
