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2 years ago

What are the missing coefficients for the skeleton equation below? cr(s) + fe(no3)2(aq) → fe(s) + cr(no3)3(aq)?

1 answer:

8 0

SThe missing coefficient for the skeleton equation below is as follows

skeleton equation

Cr(s) + Fe(No3)2(aq) ------> Fe (s) + Cr(NO3)3 (aq)
the missing coefficient are is as follows

2 Cr(s) + 3 Fe(NO3)2 ---> 3 Fe (s) + 2 Cr(NO3)3

This is obtained by making sure all the molecules are balanced in both sides

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En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu

lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0

2 years ago

A system fitted with a piston expands when it absorbs 52.7J of heat from the surroundings. The piston is working against a press

Allushta [10]

Answer: The initial volume of the system is 60.29 L.

Explanation:

According to the first law of thermodynamics,

$\Delta U = Q - W$

As it is given that heat is being added to the system so, $\Delta H$ will be positive. And, work done on the system is negative and work done by the system is positive.

So here, $\Delta U$ = -107.4 J

Q = 52.7 J

P = 0.693 atm

And, W = PdV

or, W = $P (V_{final} - V_{initial})$

So, $\Delta U = Q - P (V_{final} - V_{initial})$

-107.4 J = 52.7 J - 0.693 \times 101.325 (63.2 - V)

-160.1 = -43.79 - 70.21 (63.2 - V)

63.2 - V = $\frac{160.1 + 43.79}{70.21}$

- V = 2.90 - 63.2

V = 60.29 L

Thus, we can conclude that the initial volume of the system is 60.29 L.

6 0

2 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

2 years ago

During a process called photoact, ________ give up an electron as a part of the light-dependent reactions.

dybincka [34]

Answer:

Chloroplasts?

Explanation:

4 0

2 years ago

Consider the air over a city to be a box 100km on a side that reaches up to an altitude of 1.0 km. Clean air is blowing into the

Mariana [72]

Explanation:

It is known that equation for steady state concentration is as follows.

$C_{a} = \frac{QC}{Q + kV}$

where, Q = flow rate

k = rate constant

V = volume

C = concentration of the entering air

Formula for volume of the box is as follows.

V = $a^{2}h$

= $100 \times 100 \times 1$

= $10000 km^{3}$

Now, expression to determine the discharge is as follows.

Q = Av

= $100 \times 1 \times \frac{4 m}{s} \times \frac{km}{1000 m}$

= 0.4 $km^{3}/s$

And, m (loading) = 10kg/s,

k = 0.20/hr

as 1 $km^3 = 10^{12}$ L (if u want kg/L as concentration)

Now, calculate the concentration present inside as follows.

$C_{in} = \frac{10kg/s}{0.4 km^3/s}$

= 25 $kg/km^3$

Now, we will calculate the concentration present outwards as follows.

$C_{out} = {C_{in}}{(1 + k \times t)}$ ,

and, t = $\frac{V}{Q}$

= 25000 s or 6.94 hr

Hence, $C_{out} = \frac{25}{(1 + 0.20 \times 6.94)}$

= 10.47 $kg/km^3$

Thus, we can conclude that the the steady-state concentration if the air is assumed to be completely mixed is $C_{out} = 10.47 kg/km^3$ and $C_{in} = 25 kg/km^3$ .

4 0

2 years ago