All categories

3 years ago

How would the concentration of silver ions compare in a 1.0 x 10-16 L saturated solution to a 1.5 x 10-16 L saturated solution?

Chemistry

1 answer:

VMariaS [17]3 years ago

3 0

The second one is more concentrated as they both times with the same thing but the second one (1.5) is bigger

You might be interested in

for the reaction shown compute the theoretical yield of product in moles each of the initial quantities of reactants. 2 Mn(s)+3

Y_Kistochka [10]

Answer:

2 mole MnO₂

Explanation:

2Mn(s) + 2O₂(g) => 2MnO₂(s)

4 0

3 years ago

Read 2 more answers

“Gasoline boils at a relatively low temperature (about 150°C). The kerosene is removed at around 200°C, followed by diesel oil a

ANEK [815]

Well its Organic Chemistry... and is the Fractional Distillation of Crude Oil

3 0

3 years ago

Based on the chart, which food chain best models a flow of energy in this ecosystem?

GarryVolchara [31]

Answer:

Explanation:

8 0

3 years ago

Sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas:NaH (s) H2O (l) → NaOH (aq) H2 (g)A

Anuta_ua [19.1K]

NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)

You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)

So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.

PV = nRT

n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)

n = 0.0385 moles H2

moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required

mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH

Brainliest Please :)

7 0

3 years ago

Read 2 more answers

The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38

adell [148]

Answer:

$1.21x10^{-3}$ M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x $10^{-3}$

H = $\frac{0.21}{1.38x10^{-3} }$

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = $1.21x10^{-3}$ M

7 0

3 years ago