Mitosis has 4 steps and meiosis has 5 steps so its 9 steps total here's a photo to help you if you need it
I found these four statements for that question:
Each molecule contains four different elements.
Each molecule contains three atoms.
Each molecule contains seven different bonds.
Each molecule contains six oxygen atoms.
The last one is true. Each molecule contains six oxygen atoms.
The number to the right of O and of (NO3) ares subscripts.
The chemical formula uses subscripts to indicate the number of atoms.
The subscript 2 in (NO3)2 means that there are two NO3 radicals.
And the subscript 3 to the right of O means that each NO3 radical has three atoms of O.
Then, the number of atoms of O is 2 * 3 = 6.
So, the true statement is the last one: each molecule of Ba (NO3)2 has six atoms of O.
From that molecule you can also tell:
- Each molecule contains one atom of barium
- Each molecule contains two atoms of nitrogen
- Each molecule contains two NO3 radicals
<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ
<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. It is represented as 
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}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28MgCl_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Mg%28OH%29_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCl%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-641.8%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%281%5Ctimes%20%28-924.5%29%29%2B%282%5Ctimes%20%28-92.30%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-16.3kJ)
Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ
Answer:
190.4g
Explanation:
1.6mol of KBr (119.002g KBr/1 mol) = 190.4g
since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.
Answer:
0.00050553
Explanation:
when the power of ten is negative, move the decimal to the left
hope this helped!
