What does a solute do in a solution

2) A balloon was inflated to a volume of 5.0 liters at a temperature of

TEA [102]

Answer:

6.12 L

Explanation:

Given that,

Initial volume, V₁ = 5 L

Initial temperature, T₁ = 7.0°C = 343 K

Final temperature, T₂ = 147°C = 420 K

We need to find its new volume. The relation between volume and temperature is given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{5\times 420}{343}\\\\V_2=6.12\ L$

So, the new volume is 6.12 L.

8 0

2 years ago

Be + O2 --> BeO Balance this and what's the type of reaction?

max2010maxim [7]

Answer:

2Be + O2 = 2BeO

its a synthesis

Explanation:

7 0

2 years ago

The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval

worty [1.4K]

<h3>Answer;</h3>

1.0875 x 10-2 atm

<h3>Explanation;</h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t

The average rate of disappearance of ozone ... is found to

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t

rate = -(1/2)*(-7.25 × 10^–3 atm)

= 3.625 × 10^–3 atm

Now use the other part of the expression:

rate = +(1/3)∆[O2)∆t

3.625 × 10–3 atm = +(1/3)∆[O2)/t

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

= 1.0875 x 10-2 atm over the same time interval

4 0

3 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

3 years ago

Hello,

nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

Ferrous Sulphate $(FeSO4)$ is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide $(SO_3)$ , Sulphur Dioxide $(SO_2)$ as well as Ferric Oxide $(Fe_2O_3)$ .

We can hence, frame a skeletal equation of this reaction and try to balance it.

Hence,

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

Now,

a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of steps:

1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside $FeSO_4$ .

2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:

Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :

$2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds decompose into smaller elements and compounds. Here, as Ferrous Sulphate decomposes into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is Decomposition Reaction.

7 0

3 years ago