Answer=4
Only solids can create regular geometric patterns... CO2 is a gas, H2O is a liquid and Sodium Chloride doesn't count because it's dissolved in a solution...
Once you have constructed an effective hypothesis, the next step in the scientific inquiry process is to
test the hypothesis through experimentation. This is a great opportunity for students to start a science
notebook, if they have not yet started recording their progress.
Steps to Identifying and Conducting an Appropriate Experiment to Test a Hypothesis
1) Present Hypotheses
Make a list of all potential hypotheses to be tested.
2) Make Predictions
For each hypothesis, ask what would be true if the hypothesis were true.
3) Write the Experimental Procedure
The experimental procedure is a step-by-step recipe for the science experiment. A good
procedure contains enough detail that someone else could easily duplicate the
experiment. Once you have formed a hypothesis, you will need to develop your
experimental procedure to test whether your hypothesis is true or false.
4) Identify the Independent and Dependent Variables
The first step of designing the experimental procedure involves planning how to change
the independent variable and how to measure the impact that this change has on the
dependent variable. To guarantee a fair test when conducting the experiment, make sure
that the only thing changing is the independent variable. All controlled variables must
remain constant.
5) Design the Experiments
How can you identify an appropriate experiment that will effectively test your
hypothesis? Begin by asking yourselves, “What can I do that will give me one result if my
hypothesis is true, and a different result if my hypothesis is false?” Design at least one
possible experiment for each hypothesis. Be sure that each experiment tests only one
hypothesis.
Fixed vs Variable Oxidation is given below.
Explanation:
1.In its compounds, hydrogen has an oxidation number of +1, except. hydrides where the. oxidation number of hydrogen is -1. In their compounds, the metals with fixed oxidation states have the oxidation number that. corresponds with the fixed oxidation number.
A variable oxidation state is a value that determines the charge of the atom depending on certain conditions.
2. Oxidation state of elements is considered to be of the most important in the study of chemistry. For some elements, this figure is constant known as fixed oxidation , while for others it is variable is called variable oxidation state.
3. MgCl2 : magnesium is in Group IIA and all elements in Group IIA have fixed oxidation numbers of +2
FeCl2 : iron has a variable oxidation number of either +2 or +3 and is not fixed
<span>You need to have NAD+ as a source of oxidation for the pyruvate, as well as a supply of coenzyme A. CO2 is released by the pyruvate as a carboxyl group is removed</span>
<u>Answer:</u> The 
for HCN (g) in the reaction is 135.1 kJ/mol.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ](https://tex.z-dn.net/?f=-870.8%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28-80.3%29%29%2B%283%5Ctimes%20%280%29%29%2B%282%5Ctimes%20%28-74.6%29%29%5D%5C%5C%5C%5C%5CDelta%20H_f_%7B%28HCN%29%7D%3D135.1kJ)
Hence, the for HCN (g) in the reaction is 135.1 kJ/mol.
