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sweet-ann [11.9K]
3 years ago
6

Which solution will form a precipitate when mixed with a solution of aqueous na2co3? which solution will form a precipitate when

mixed with a solution of aqueous ? nh4cl(aq) k2so4(aq) nabr(aq) cucl2(aq)?
Chemistry
1 answer:
trapecia [35]3 years ago
4 0

Answer:

CuCl_{2}(aq.) will form a precipitate of insoluble CuCO_{3} when aqueous Na_{2}CO_{3} is added.

Explanation:

According to solubility rule-

all carbonates are insoluble except group IA compounds and NH_{4}^{+}

all salts of sodium are soluble

When Na_{2}CO_{3} is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of Na_{2}CO_{3} into CuCl_{2}(aq.) will result precipitation of insoluble CuCO_{3}

Reaction: Na_{2}CO_{3}(aq.)+CuCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CuCO_{3}(aq.)

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Caculate the mass of oxygen needed to burn 175.0grams of octane
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A water wave traveling in a straight line on a lake is described by the equation
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Calculate the density of the aluminum cylinder with a diameter 0f 1.3 cm weighing 18 grams. Height of the cylinder is 5.2 Cm. Fi
Masja [62]

Answer:

Percent error = 3.7%

Explanation:

Given data:

Density of Al cylinder = ?

Weight of cylinder = 18 g

Diameter = 1.3 cm

Height = 5.2 cm

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Percent error = ?

Solution:

First of all we will calculate the volume of cylinder through given formula.

V = πr²h

r = diameter /2

V = 22/7 × (0.65 cm)²× 5.2 cm

V = 22/7 × 0.4225cm²× 5.2 cm

V = 6.89 cm³

Now we will calculate the density.

d = m/v

d = 18 g/ 6.89 cm³

d = 2.6 g/cm³

Percent error:

Percent error = measured value - actual value /actual value × 100

Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³  × 100

Percent error = 3.7%

Negative sign shows that measured or experimental value is less than actual value.

6 0
3 years ago
A possible mechanism for the overall reaction Br2 (g) + 2NO (g) → 2NOBr (g) is NO (g) + Br2 (g) rightwards arrow with k space 1
omeli [17]

Answer : The rate law for formation of NOBr based on this mechanism is, \frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2

Explanation :

The overall reaction is:

Br_2(g)+2NO(g)\rightleftharpoons 2NOBr(g)

Rate law = k[Br_2][NO]^2

The first step of the overall reaction is:

NO(g)+Br_2(g)\overset{k_1}{\rightarrow} NOBr_2(g)

NO(g)+Br_2(g)\overset{k_1^-}{\leftarrow} NOBr_2(g)

Rate law 1 = k_1[Br_2][NO]

Rate law 2 = k_1^-[NOBr_2]

The second step of the overall reaction is:

NOBr_2(g)+NO(g)\overset{k_2}{\rightarrow} 2NOBr

Rate law 3 = k_2[NOBr_2][NO]

Now rate law of overall reaction can be obtained as follows.

We are multiplying rate law 1 and rate law 3 and dividing by rate law 2, we get:

Rate law = \frac{[k_1[Br_2][NO]]\times [k_2[NOBr_2][NO]]}{[k_1^-[NOBr_2]]}

Rate law = \frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2

Rate law = k[Br_2][NO]^2

The rate law for formation of NOBr based on this mechanism is, \frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2

4 0
3 years ago
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