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blagie [28]
3 years ago
7

If equal amounts of silver ion are added to silver bromide and silver chloride, why does the bromide mixture precipitate first?

(Ksp of silver bromide = 5.3 mc022-1.jpg 10-13 and Ksp of silver chloride = 1.8 mc022-2.jpg 10-10)
Chemistry
1 answer:
VARVARA [1.3K]3 years ago
5 0
Answer is: precipitation requires fewer Ag⁺<span> ions in AgBr than in AgCl.
Chemical reactions:
Ksp(KBr) = 5,3</span>·10⁻¹³.
Ksp(KCl) = 1,8·10⁻¹⁰.
Ksp is <span>solubility product constant. The higher the Ksp value, substance is more soluble. KBr has lower Ksp, so it is easier to form precipitant of KBr than KCl.</span>
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Finding the pH for [H+] = 9.4 * 10-3 M?
saul85 [17]

Answer:

pH = 2.0

Explanation:

To find the pH of a solution, take the -log[H+]. In this case, the -log(9.4 x 10^-3) equals 2.02687 which makes 2.0 when accounting for significant figures.

3 0
2 years ago
Which of the following shows three elements in order of decreasing electronegativity?
damaskus [11]

Answer:

The correct answer is option D which is the decreasing order of conductivity is Mn, O, Ge.

Explanation:

You can easily answer this if you know the periodic trends. For the property of electrical conductivity, it decreases across a period and decreases also down a group. Thus, the most conductive element must be Mn, while the least conductive one is Ge. So, the answer is: -Mn, O, Ge

7 0
3 years ago
Purchasing the correctly sized BMX bike is based on the height of the rider. In order to fit a customer, the salesman can use th
Phantasy [73]
Answer: Your best answer is A.

Explanation: 1.35 is a constant and thus is does not change. The equation can be simplified to-
“b = (0.29 x h)+ 1.35”
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“b = (0.29 x 60)+ 1.35”
Then to-
“b = (17.4) + 1.35” or “b = 17.4 + 1.35”
And therefore the size of the BMX bike frame in inches would be-
“b = 18.75”

Hope this helped :)
5 0
2 years ago
Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---&gt; CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ  

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

_________________________________

The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


7 0
2 years ago
Fill in the missing data point. Show all calculations leading to an answer.
Aleksandr-060686 [28]

Answer:  

1090 mmHg  

Explanation:  

We know that with gases we must use a Kelvin temperatures, so let’s try a plot of pressure against the Kelvin temperature.  

We can create a table as follows  

<u>t/°C</u>  <u>T/K</u>  <u>p/mmHg</u>  

  10   283      726  

  20  293      750  

  40   313      800  

  70  343      880  

100  373       960  

150  423        ???  

I plotted the data and got the graph in the figure below.  

It appears that pressure is a linear function of the Kelvin temperature.  

y = mx + b  

where x is the slope and b is the y-intercept.

===============

<em>Calculate the slope  </em>

I will use the points (275, 700) and (380, 975).  

Slope = Δy/Δx = (y₂ - y₁)/(x₂ -x₁) = (975 -700)/(380 – 275) = 275/105 = 2.619  

So,  

y = 2.619x + b  

===============

<em>Calculate the intercept </em>

When x = 275, y = 700.  

700 = 2.619 × 275 + b  

700 = 720 + b     Subtract 720 from each side and transpose.  

b = -20  

So, the equation of the graph is  

y = 2.619x -20  

===============

<em>Calculate the pressure</em> at 423 K (150°C)  

y = 2.619 × 423 - 20  

y = 1110 - 20  

y = 1090

At 150 °C, the pressure 1090 mmHg.  

The point is approximately at the position of the black dot in the graph.  

7 0
3 years ago
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