Answer:
1. Hidracidas a. MX
2 Acidas c. MHXO
3. Oxacidas b. MXO
4. Basicas d. M(OH)X
Explanation:
¡Hola!
En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.
Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:
1. Hidracidas a. MX
2 Acidas c. MHXO
3. Oxacidas b. MXO
4. Basicas d. M(OH)XO
En las que M se refiere a un metal, X a un no metal, H a hidrógeno y O a oxígeno.
¡Saludos!
To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent.
So we get:
So you get about:
So, the percent composition of Hydrogen in NH4HCO3 is 6.3%
Hope this Helps! :D
According to Henry's law, solubility of solution is directly proportional to partial pressure thus,
Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.
Putting the values in equation:
On rearranging,
Therefore, solubility will be 1.88 mg of 
gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.
The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months
To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Amount remaining (N) = 5%
Original amount (N₀) = 100%
<h3>Number of half-lives (n) =?</h3>
N₀ × 2ⁿ = N
5 × 2ⁿ = 100
2ⁿ = 100/5
2ⁿ = 20
Take the log of both side
Log 2ⁿ = log 20
nlog 2 = log 20
Divide both side by log 2
n = log 20 / log 2
<h3>n = 4.32</h3>
Thus, 4.32 half-lives gas elapsed.
Finally, we shall determine the half-life of the element. This can be obtained as follow.
Number of half-lives (n) = 4.32
Time (t) = 500 years
<h3>Half-life (t½) =? </h3>
t½ = t / n
t½ = 500 / 4.32
t½ = 115.74 years
Multiply by 12 to express in months
t½ = 115.74 × 12
<h3>t½ ≈ 1389 months </h3>
Therefore, the half-life of the radioactive element in months is approximately 1389 months
Learn more: brainly.com/question/24868345
