When salt water is added to onion cells, then the cells will lose water due to osmosis, this can be observed.
Answer:
Se requerirán 14.57 gramos de MgSO₄·7H₂O, que se disolverían en 105.43 gramos de agua.
Explanation:
Si tenemos 120 gramos de una solución al 10% de sulfato de magnesio en peso, habrán en la solución (120*10/100) 12 gramos de sulfato de magnesio (MgSO₄).
Sin embargo, el reactivo que está disponible es heptahidratado (MgSO₄·7H₂O), por lo que hay que calcular <em>cuántos gramos de sulfato de magnesio heptahidratado contendrán 12 gramos de MgSO₄</em>.
<u>Calculamos las moles de 12 gramos de MgSO₄</u>, usando su masa molecular:
- 12 g MgSO₄ ÷ 120.305 g/mol = 0.0997 mol MgSO₄.
<u>Después calculamos la masa de MgSO₄·7H₂O que contendrá 0.0997 mol MgSO₄</u>, usando la masa molecular de MgSO₄·7H₂O:
- 0.0997 mol * 246.305 g/mol = 14.57 g MgSO₄·7H₂O
Para saber la cantidad de agua en la que se disolverá el reactivo, restamos la masa de soluto de la masa total de la solución:
- 120 g - 14.57 g = 105.43 g
Answer:
B
Explanation:
Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation
PV = nRT
n = PV/RT
The parameters have the following values according to the question:
P = 780mmHg, we convert this to pascal.
760mHG = 101325pa
780mmHg = xpa
x = (780 * 101325)/760 = 103,991 Pa
V= 400ml = 0.4L
T = 135C = 135 + 273.15 = 408.15K
n = ?
R = 8314.463LPa/K.mol
Substituting these values into the equation yields the following:
n = (103991 * 0.4)/(8314.463 * 408.15)
= 0.012 moles
Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules
Molar mass
H2S = 34.0 g/mol
O2 = 31.99 g/mol
S8 = 256.52 g/mol
Identifying excess reagent and the limiting of the reaction :
8 H2S(g) + 4 O2(g) = S8(I) + 8 H2O(g)
8 x 34 g H2S --------> 256. 52 g S8
35.0 g ----------------> ??
35.0 x 256.52 / 8 x 34 =
8978.2 / 272 => 33.00 g of S8
H2S is the limiting reactant
---------------------------------------------
4 x 31.99 g O2 --------------- 256.52 g S8
40.0 g O2 --------------------- ??
40.0 x 256.52 / 4 x 31.99 =
10260.8 / 127.96 = 80.16 g of S8
O2 is the excess reagent is the excess <span>reagent
</span>
------------------------------------------------------------
H2S is the limiting reactant, one that is fully consumed, it is he who determines the mass of S8 produced
33.0 g ----------- 100%
?? g ------------- 95 %
95 x 33.00 / 100 => 31.35 g
hope this helps!
Answer:
A:increase from left to right and top to bottom
Explanation:
on the periodic table it's shows it increasing for left to right and decreasing from top to bottom
