PH= −log
10
[H
+
]
= −log
10
(0.001)
= −log
10
(10
−3
)
= −(−3)log
10
10
pH=3.
01
Answer:
2,14 g / ml
Explanation:
Sabemos que el volumen de una sustancia es igual al cambio de volumen del agua cuando el objeto en cuestión se sumerge en el agua.
Dado que el volumen original del agua = 50 ml
Volumen de agua después de sumergir el objeto = 120 ml
Masa del objeto = 150 g
Ahora,
Densidad = masa / volumen
Densidad = 150g / 120-50 ml
Densidad = 150/70 ml
Densidad = 2,14 g / ml
Answer:
66s^-1 will be 1/66
then to convert to minute you multiply by 69
1/66 x 60 = 3960 mins
Answer:
3 g/mL
Explanation:
We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).
Formula
D=m/v
Given data:
Mass= 45 g
Volume= 15 mL
Now we will put the values in formula:
D=45 g/ 15 mL= 3 g/mL
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
