Answer:
Cytoplasm is the space between cell membrane and nuclear membrane filled with a translucent liquid . It contains carbohydrates , proteins , lipids , nucleic acids , sodium and pottassium salts , water and enzymes .
1) Carbon-13:
Proton-6 Neutron-7 Electron-6
2)Atomic mass of element X:
(55*10+56*20+57*70)/100=56.6
Answer:
Making oxygen
Oxygen can be made from hydrogen peroxide, which decomposes slowly to form water and oxygen:
hydrogen peroxide → water + oxygen
2H2O2(aq) → 2H2O(l) + O2(g)
The rate of reaction can be increased using a catalyst, manganese(IV) oxide. When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off.
Apparatus arranged to measure the volume of gas in a reaction. Reaction mixture is in a flask and gas travels out through a pipe in the top and down into a trough of water. It then bubbles up through a beehive shelf into an upturned glass jar filled with water. The gas collects at the top of the jar, forcing water out into the trough below.
To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water. As the oxygen collects in the top of the gas jar, it pushes the water out.
Instead of the gas jar and water bath, a gas syringe could be used to collect the oxygen.
The Answer is A
This is because sodium needs to give off one electron to be stable, form these options the only element that needs to take one electron to be stable is fluorine thus A is the answer
Answer:
91383 J
Explanation:
The equation of the reaction can be represented as:
------>
Given that:
The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.
The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.
= 
where:
= enthalpy of reaction
= the difference in the heat capacities of the products and the reactants.
∴
= 
= ![1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'](https://tex.z-dn.net/?f=1%2891300%20J.mol%5E%7B-1%7D%20%29%20%2B%5Cint%5Climits%5E%7B435%7D_%7B298.15%7D%20%5B%7B%2829.86%29-%5Cfrac%7B1%7D%7B2%7D%2829.38%29-%5Cfrac%7B1%7D%7B2%7D29.13%7D%5DJ.K%5E%7B-1%7D.mol%5E%7B-1%7D%20%5C%2C%20dT%27)
= 91300 J + (0.605 J.K⁻¹)(435-298.15)K
= 91382.79 J
≅ 91383 J
