The suggestion is to prevent a puddle of the liquid present in the sample from forming or from it leaking on to the surface on which it is placed. For example, if precipitates of a solid are removed from water and then placed on filter paper to dry, the water will soak into the filter paper and then leak on to the counter on which it is placed. If this precipitate were placed in a watch glass or weighing paper, the water would only evaporate and would not contaminate the sample.
The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²
Given, that a space shuttle requires a 20.7 cm² patch
We have to convert the patch's area from cm² into km².
Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.
Patch area of the space shuttle is 20.7 cm²
1 cm = 0.00001 km
or, 1 cm² = (0.00001 km)²
or, 1 cm² = 10⁻¹⁰km²
20.7 cm² = 20.7 × 10⁻¹⁰km²
20.7 cm² = 2.07 × 10⁻⁹ km²
The patch area in square kilometers is 2.07 × 10⁻⁹ km²
To learn more about unit conversion, visit: brainly.com/question/11543684
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The answer is 615.91 grams of <span>n2f4
Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams
For an non spontaneous reaction between silver (Ag) and copper (Cu) and their ions, Cu is the oxidizing agent while Ag+ is the reducing agent,
The following reactions will take place;
Anode Cu = Cu+2 + 2e- E= +0.34 volts
Cathode; Ag+ + e = Ag E = +0.80 volts
The net reaction will be Cu + 2Ag+ = Cu+2 + 2Ag
Thus, the voltage will be
= +0.80 - (+0.34)
