12) Which model of the atom is thought to be true?

The volume of oxygen, collected over water, is 185 mL at 25 degrees Celsius and 600 torr. calculate the dry volume of the oxygen

ivolga24 [154]

Answer:

0.1593 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

P₁V₁T₂ = P₂V₂T₁

P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.

P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.

∴ V₂ = P₁V₁T₂/P₂T₁ = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = 0.1593 L.

4 0

3 years ago

Is BH2F a gas or a liquid and what is the boiling point

hram777 [196]

Hey I tried but could not find your answer, but maybe you could try to calculate the average of bpts of the different elements in the compound to find the boiling point. And when you know the bpt, you can find out whether it is a liquid or gas. But after searching a lot, I found it was a solid. Good luck!

7 0

3 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago

Heredity is defined as

AnnyKZ [126]

Answer:

B.

the passage of genetic instructions from one generation to the next generation.

These are called genes. One mate reporduces with another made and the genetic buildup merges 50 % and 50% with genetics(if it's meiosis), or DNA codes from the parents to the offspring that then possess some of the genes. That shows heridity.

8 0

3 years ago