Answer:
1 tell your family difrent ways to safe water
2 dont leave the water running when you brush your teeth
3 dont dump cups of water out for no reason
Explanation:
Answer:
Lithium
Explanation:
The answer is Lithium because the Alkali metals family is in the group 1A and the element neon is in the second period. If you look at the periodic table group 1A and the second period connects at the element Lithium.
Answer:
17.65 grams of O2 are needed for a complete reaction.
Explanation:
You know the reaction:
4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O
First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:
- N: 14 g/mol
- H: 1 g/mol
- O: 16 g/mol
So, the molar mass of the compounds in the reaction is:
- NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
- O₂: 2*16 g/mol= 32 g/mol
- NO: 14 g/mol + 16 g/mol= 30 g/mol
- H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol
By stoichiometry, they react and occur in moles:
- NH₃: 4 moles
- O₂: 5 moles
- NO: 4 moles
- H₂O: 6 moles
Then in mass, by stoichiomatry they react and occur:
- NH₃: 4 moles*17 g/mol= 68 g
- O₂: 5 moles*32 g/mol= 160 g
- NO: 4 moles*30 g/mol= 120 g
- H₂O: 6 moles*18 g/mol= 108 g
Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O₂≅17.65 g
<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>
Answer: 3M
Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.
Formula used :
where,
n = moles of solute
= 0.6 moles
= volume of solution in ml= 200 ml
Now put all the given values in the formula of molarity, we get
Therefore, the molarity of solution will be 3M.
Answer:
Explanation:
To find the concentration; let's first compute the average density and the average atomic weight.
For the average density
; we have:

The average atomic weight is:

So; in terms of vanadium, the Concentration of iron is:

From a unit cell volume 

where;
= number of Avogadro constant.
SO; replacing
with
;
with
;
with
and
with 
Then:
![a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2FA_%7BFe%7D%20%5D%20%2B%20%5BC_v%2FA_v%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2F%5Crho_%7BFe%7D%20%5D%20%2B%20%5BC_v%2F%5Crho_v%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100-C_v%29A_%7Bv%7D%20%5D%20%2B%20%5BC_v%2FA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100-C_v%29%2F%5Crho_%7Bv%7D%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100A_%7Bv%7D-C_vA_%7Bv%7D%29%20%5D%20%2B%20%5BC_vA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100%5Crho_%7Bv%7D%20-%20C_v%20%5Crho_%7Bv%7D%29%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
Replacing the values; we have:



