1A: The legs can be a adjusted, as well as the sand can be swapped out. It’s a very good design for running multiple tests.
1B: He could add books or something under the front or back legs in order to increase/decrease the incline, therefore imitating the hypothesis.
1C: He can change out the sand grains to finer ones, or coarser ones, and record his results of each test.
2: If he sets the model at a steep incline and tests it with coarse sand and fine sand, seeing which one makes a narrower, deeper hole.
Answer:
B. Positive charge is caused by lack of electrons. A positive ion is formed by the loss of negatively charged electrons. Although the number of protons does not change in the ion, there is an excess number of protons over electrons which produces the positive charge. All electrons in the outer energy level are lost.
C. If there are more electrons than protons, then the element is a negative ion. The amount of neutrons does not play a factor into making a difference between an atom or an ion.
Explanation:
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Answer:
A.) ![K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
Explanation:
The general Kb expression is:
![K_b = \frac{[HA][OH^-]}{[A^-]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BHA%5D%5BOH%5E-%5D%7D%7B%5BA%5E-%5D%7D)
In this equation
-----> Kb = equilibrium constant
-----> [HA] = acid
-----> [A⁻] = base
Since liquids are not included in equilibrium expressions, H₂O should not be present. The products are in the numerator while the reactant are in the denominator. In this reaction, CH₃NH₂ is acting as a base and CH₃NH₃⁺ is acting as an acid.
As such, the expression is:
substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span>
Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span>
<span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span>
<span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>
