Answer:
Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.
Explanation:
Hola,
En este caso, la reacción llevada a cabo es:
De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:
Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:
Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:
Esto quiere decir que de 50 gramos de oxígeno, solo 20 gramos reaccionan para formar 70 gramos de óxido de calcio.
Saludos!
Answer:
All of them affect the DNA
Explanation:
Chemicals are compounds that can pass through cell membranes and modificate the DNA, elevated temperatures can denaturalize the cell and therefore damage the DNA, ionizing radiation can pass through cell organelles and reach the nucleus affecting the DNA, and viruses inject its DNA into the genome and modify it.
Answer:
v = 37.9 ml
Explanation:
Given data:
Mass of compound = 1.56 kg
Density = 41.2 g/ml
Volume of compound = ?
Solution:
First of all we will convert the mass into g.
1.56 ×1000 = 1560 g
Formula:
D=m/v
D= density
m=mass
V=volume
v = m/d
v = 1560 g / 41.2 g/ml
v = 37.9 ml
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer: D=8.27 g/cm³
Explanation:
Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.
With this equation, the m³ and kg cancel out, and we are left with g/cm³.
D=8.27 g/cm³
