Determine the volume of fluid in the graduated cylinder shown. 22.5 mL 22 mL 23.0 mL 23 mL

What is a mineral used in making fertillizer 1. silver 2. silica 3. tin 4. gypsum

Marat540 [252]

The Answer Is Gypsum

Hope That Helps!

5 0

3 years ago

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En la reacción I2(g) + Br2(g) « 2 IBr(g), Keq = 280 a 150°C. Suponga que se permite que 0.500 mol de IBr en un matraz de 1.00 L

love history [14]

Answer:

P IBr: 15.454atm

I₂: 0.923 atm

P Br₂: 0.923atm

Explanation:

Basados en la reacción:

I₂(g) + Br₂(g) ⇄ 2 IBr(g)

La constante de equilibrio, Keq, es definida como:

$Keq = \frac{P_{IBr}^2}{P_{I_2}P_{Br_2}}$

Se cumple la relación de Keq = 280 cuando las presiones están en equilibrio

Usando PV = nRT, la presión inicial de IBr es:

P = nRT / V; 0.500mol*0.082atmL/molK*423.15K / 1.00L = 17.3 atm

Siendo las presiones en equilibrio:

P IBr: 17.3 - 2X

P I₂: X

P Br₂: X

Donde X representa el avance de reacción.

Remplazando en Keq:

280 = (17.3 - 2X)² / X²

280X² = 4X² - 69.2X + 299.29

0 = -276X² - 69.2X + 299.29

Resolviendo para X:

X = -1.174 → Solución falsa. No existen presiones negativas

X = 0.923 → Solución real

Así, las presiones parciales en equilibrio de cada compuesto son:

P IBr: 17.3 - 2X = 15.454atm

P I₂: X = 0.923atm

P Br₂: X = 0.923atm

3 0

3 years ago

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Water is a produced when 30.0 grams of hydrogen reacts with 80.0 grams of oxygen what is the limiting reagent

irakobra [83]

The balanced equation for the reaction is as follows
2H₂ + O₂ --> 2H₂O
stoichiometry of H₂ to O₂ is 2:1
number of H₂ moles - 30.0 g / 2 g/mol = 15 mol
number of O₂ moles - 80.0 g / 32 g/mol = 2.5 mol
limiting reactant is the reagent in which only a fraction is used up in the reaction
if H₂ is the limiting reactant
if 2 mol of H₂ requires 1 mol of O₂
then 15 mol of H₂ requires 1/2 x 15.0 = 7.5 mol of O₂
but only 2.5 mol of O₂ is required
this means that O₂ is the limiting reagentt and H₂ is in excess

7 0

3 years ago

The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp

Eva8 [605]

Answer:

$[A]_0=0.400M$

Explanation:

Hello.

In this case, since the first-order reaction is said to be linearly related to the rate of reaction:

$r=-k[A]$

Whereas [A] is the concentration of hydrogen peroxide, when writing it as a differential equation we have:

$\frac{d[A]}{dt} =-k[A]$

Which integrated is:

$ln(\frac{[A]}{[A]_0} )=-kt$

And we can calculate the initial concentration of the hydrogen peroxide as follows:

$[A]_0=\frac{[A]}{exp(-kt)}$

Thus, for the given data, we obtain:

$[A]_0=\frac{0.321M}{exp(-2.54x10^{-4}s^{-1}*855s)}$

$[A]_0=0.400M$

Best regards!

3 0

3 years ago

What is the partial pressure of argon, par, in the flask? express your answer to three significant figures and include the appro

monitta

Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C°
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n = 1.45/mass weight of argon
= 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm

3 0

3 years ago