Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer:
-32 Fahrenheit converts to 237.594 Kelvin
Answer:
sample A
Explanation:
the first one because of the ppm value
Answer:
2.25×10¯³ mm.
Explanation:
From the question given above, we obtained the following information:
Diameter in micrometer = 2.25 μm
Diameter in millimetre (mm) =?
Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m
2.25 μm = 2.25×10¯⁶ m
Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:
1 m = 1000 mm
Therefore,
2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm
2.25×10¯⁶ m = 2.25×10¯³ mm
Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.
Answer:
See explanation
Explanation:
In putting up prospective structures for the compounds, we must consider the tetra valency of carbon. Again you must read the description carefully before drawing each structure.
In each of the proposed structures, the number of atoms present must reflect the condensed structural formula given in the question.
Proposed structures for each chemical specie in the question are shown in the images attached.
