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lesya [120]
2 years ago
5

How many liters of a 0.225 M solution of KI are needed to contain 0.935 moles of KI?

Chemistry
1 answer:
Katyanochek1 [597]2 years ago
6 0

Answer:

4.16L

Explanation:

From the question given, we obtained the following data:

Molarity = 0.225 M

Number of mole of KI = 0.935mole

Volume =?

Molarity = mole / Volume

Volume = mole /Molarity

Volume = 0.935/0.225

Volume = 4.16L

Therefore, 4.16L of KI is needed.

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Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2Mg (s) + O2 (g) →→ 2MgO (s) When 4.50 g of magnes
Klio2033 [76]

Answer:

7.46 g

Explanation:

From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.

The mole ratio is 1:1

mole = mass/molar mass

mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole

0.185 mole Mg will tiled 0.185 MgO

Hence, theoretical yield of MgO in g

mass = mole x molar mass

            0.185 x 40.3 = 7.46 g

6 0
2 years ago
Read 2 more answers
When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. When aqueous so
MakcuM [25]

Answer:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

Explanation:

We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.

Aqueous AgNO₃ and KI will dissociate in solution as follow:

AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)

KI (aq) —> K⁺(aq) + I¯(aq)

Aqueous AgNO₃ and KI will react as follow:

AgNO₃ (aq) + KI (aq) —>

Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)

Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.

Thus, the net ionic equation is:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

3 0
2 years ago
Determine the transition length at the entrance to a 10 mm tube through which 100 percent glycerol at 60 C is flowing at a veloc
Readme [11.4K]

Answer: Option (d) is the correct answer.

Explanation:

The given data is as follows.

Tube diameter d = 10 mm = 0.01 m

Velocity of glycerol, v = 0.5 m/s

Density of glycerol (\rho) = 1240 kg/m3

Dynamic viscosity of glycerol (\mu) = 0.0813 pa.s

Reynolds number (Re) = \rho \times velocity \times \frac{density}{\mu}

                                     = 1240 \times 0.5 \times \frac{0.01}{0.0813}

                                     = 76.26

Therefore, according to Reynolds number we can say that flow is laminar.

                     Lt = 0.05 \times Re \times d

                         = 0.05 \times 76.26 \times 0.01

                         = 0.03813 m

As it is known that 1 m = 1000 mm. Hence, in 0.03813 m will be equal to 0.03813 m \times \frac{1000 mm}{1 m}

                         = 38.13 mm

Thus, we can conclude that the transition length of glycerol is 38.13 mm.

5 0
2 years ago
How do models help scientists predict the polarity of molecules?
yuradex [85]

Answer:There are three main properties of chemical bonds that must be considered—namely, their strength, length, and polarity. The polarity of a bond is the distribution of electrical charge over the atoms joined by the bond. Specifically, it is found that, while bonds between identical atoms (as in H2) are electrically uniform in the sense that both hydrogen atoms are electrically neutral, bonds between atoms of different elements are electrically inequivalent. In hydrogen chloride, for example, the hydrogen atom is slightly positively charged whereas the chlorine atom is slightly negatively charged. The slight electrical charges on dissimilar atoms are called partial charges, and the presence of partial charges signifies the occurrence of a polar bond.

Explanation:

8 0
2 years ago
Read 2 more answers
Iodine-131 has a half-life of 8.10 days. In how many days will 50 grams of Iodine-131 decay to one-eighth of its original amount
shtirl [24]
What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half 
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25                                               8.1
25/2 =12.5                                        +  8.1
12.5/2= 6.25                                      +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8  of its original mass.

(good luck on the regent if thats what your studying for :)

5 0
2 years ago
Read 2 more answers
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