What is the molar mass of Fe2Br3?

A sample of oxygen gas occupies 3.60 liters at a pressure of 1.00 atm. If temperature is held constant, what will be the volume

Soloha48 [4]

Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L

At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then

PV = c

Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.

It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,

P₁V₁ = P₂V₂

Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.

In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,

1.00 atm x 3.60 L = 2.50 atm x y

We cleared y to find V₂,

V₂ = y = $\frac{1.00 atm x 3.60 L}{2.50 atm}$ = 1.44 L

Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>

3 0

3 years ago

Approximately 1 mL of two clear, colorless solutions-0.1 M Ba(NO3)2 and 0.1 M Na2SO4- were combined. Upon mixing, a thick, milky

murzikaleks [220]

Answer: balanced chemical equation: $Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Net ionic equation : $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical equation is:

$Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

$Ba^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2Na^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are sodium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

4 0

3 years ago

I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:

dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios. This makes up for using the units that wouldn't appropriate elsewhere.

2. Use the equation P₁V₁ = P₂V₂. Solve for V₂.

(3.05 L)(870 kPa) = (969 kPa)(V₂)

V₂ = 2.74 L.

3. Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.

(3.32 L)/(360 K) = (4.50 L)/(T₂)

T₂ = 488 K.

4. Do the same as above, but for V₂.

(5.10 L)/(-56°C) = V₂/(-82°C)

V₂ = 7.47 L

5. Use the equation V₁/n₁ = V₂/n₂. Solve for V₂.

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)

V₂ = 38.6 L

6. Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)

n₂ = 2.85 mol

7. Use the equation P₁/T₁ = P₂/T₂. Solve for T₂.

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)

T₂ = 319 K

8. Do the same as above, but for P₂.

In this specific case, however you will need to convert units. Since both temperatures don't have the same sign, the ratio won't come out right. Convert to Kelvin. Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K 25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

6 0

3 years ago

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

Taya2010 [7]

Answer:

Here's what I get

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

7 0

3 years ago

Question 5(Multiple Choice Worth 5 points)

mars1129 [50]

Answer:

Density

Explanation:

8 0

2 years ago

Read 2 more answers