What volume will 25.0g ammonia occupy at stp

How many moles of each substance are needed to prepare the following solutions?

Afina-wow [57]

A) 7.5 % m / V

7.5 g -------- 100 mL ( solution )
mass g ------ 45.0 mL

mass g = 45.0 x 7.5 / 100

mass g = 337.5 / 100 => 3.375 g

1 mole KCl -------------- 74.55 g
moles KCl ---------------- 3.375 g

moles KCl = 3.375 x 1 / 74.55 => 0.0452 moles of KCl
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b) 9.5 % ( m / V ) :

9.5 g -------- 100 mL ( solution )
mass g ------ 400.0 mL

mass g = 400.0 x 9.5 / 100

mass g = 3800 / 100 => 38 g

1 mole acetic acid -------------- 60.05 g
moles acetid acid ---------------- 38 g

moles acetid acid = 38 x 1 / 60.05 => 0.6328 moles of acetid acid
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hope this helps!

4 0

3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

Plzzzzzzzzzzzzzzzz help

lyudmila [28]

Cells correct error in their DNA by three processes: proofreading (errors that occur during replication are rectified), mismatch repair (the mismatched base pairs are corrected just after the DNA replication), and damage repair mechanisms (including direct reversal by cell enzymes, excision repair by replacement or removal of the wrong base, and double stranded base repair by repairing the breaks in the DNA by homologous or non-homologous recombination methods). Pick according to this answer >)

3 0

4 years ago

Two concentration cells are prepared, both with 90.0 mL of 0.0100 M Cu(NO₃)₂ and a Cu bar in each half-cell. (b) Calculate Ecell

pogonyaev

The Ecell when an additional 10.0 mL of 0.500 M NH₃ is added is 0.156 V.

When NH3 is added to the first cell, Nh3 react with Cu(NO3) react to form complex.

Thus, Cu2+ ion concentration decrease in the first cell.

Anode

Cu ---- Cu(2+) + 2e-

Cathode

Cu(2+) + 2e- ------ Cu

Ecell can be calculated as

Ecell = E°cell - (0.059/2) log{ Cu(2+) / Cu(+2) cathode}

[Cu2+] cathode = 90ml × 0.01M = 9 × 10^(-4) moles

or,

0.129 = 0 - (0.059/2) log ( Cu(+2) / 9 × 10^(-4))

[Cu(2+) ] anode = 3.8 × 10^(-8) mol

<h3>Chemical reaction of Nh3 with Cu2+</h3>

(Cu2+) + 4 NH3 -----; Cu(NH3)4(2+)

Kf can be given as

Kf = [Cu(NH3)4(2+)]/ [Cu2+] [ NH3]^4

Concentration of NH3 = 19 ml × 0.5 M

= 0.005 m

Kf = 0.005/ (3.8 × 10^(-8) mol) × (0.005) ^4

= 2.09 × 10^14.

If 10ml NH3 id added in the solution, then the total concentration of NH3 can be 20ml and 0.5 M = 0.01mol

Now, we can calculate the [Cu2+] anode

[Cu2+] anode = [Cu(NH3)4(2+)]/ Kf × [ NH3]^4

By substituting all the values, we get

= 4.78 × 10^(-9) moles.

E cell = E°cell - (0.059/2) log{ Cu(2+) / Cu(+2)

0- (0.059/2) log{ 4.78 × 10^(-9) / 9 × 10^(-4))

E cell = 0.156 V.

Thus, we calculated that the Ecell when an additional 10.0 mL of 0.500 M NH₃ is added is 0.156 V.

learn more about Ecell:

brainly.com/question/861659

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7 0

2 years ago

What is a naturally occurring solid, inorganic compound or element?

Oksana_A [137]

Answer is: naturally occurring solid, inorganic compound or element is mineral.
Minerals usually have crystalline form, that means than crystal constituents (atoms, molecules or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.
For example magnetite is a mineral of iron oxide.

3 0

3 years ago