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klio [65]
3 years ago
8

Why cant you ice skate on a lake when it is not frozen​

Chemistry
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

You can't ice skate on a liquid, when it is frozen it is a solid, when it's unfrozen it is a liquid

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Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 2
erastovalidia [21]

Answer:

Chloroform= limiting reactant

0.209mol of CCl4 is formed

And 32.186g of CCl4 is formed

Explanation:

The equation of reaction

CHCl3 + Cl2= CCl4 + HCl

From the equation 1 mol of

CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4

From the question

25g of CHCl3 really with Cl2

Molar mass of CHCl3= 119.5

Molar mass of Cl2 = 71

Hence moles of CHCl3= 25/119.5 = 0.209mol

Moles of Cl2 = 25/71 = 0.352mol

Hence CHCl3 is the limiting reactant

Since 1 mole of CHCl3 gave 1mol of CCl4

It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4

Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g

6 0
3 years ago
Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100
vesna_86 [32]

<u>Answer:</u> The solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}

                      3s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2

We are given:

K_{sp}=2.5\times 10^{-33}

Putting values in above equation, we get:

2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L

Hence, the solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

6 0
3 years ago
Use dimensional analysis to convert 14.5mi/hr to km/s
Liula [17]

Answer:

0.006 48 km/s

Explanation:

1. Convert miles to kilometres

14.5 mi × (1.609 km/1 mi) = 23.33 km

2. Convert hours to seconds

1 h × (60 min/1h) × (60 s/1 min) = 3600 s

3. Divide the distance by the time

14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s

5 0
3 years ago
Which of the following terms describes an accumulation of rocky, sandy, or clayey material deposited at the end of a glacier?
OLEGan [10]

Answer:

Option (D)

Explanation:

<u>Eskers are the long ridges that are comprised of rocks, sands and clay particles and are deposited towards the end of the glaciers</u>. These are fluvioglacial depositional features. These particles are exposed after the glaciers recede. These ridges are formed parallel to the earlier flow direction of ice. The size of eskers is generally smaller as it carries smaller particles such as rocks, sands, and gravels, in comparison to the different type of moraines. It is because the flow velocity decreases as the glaciers melt. So, these eskers are formed at the end of the glaciers.

Thus, the correct answer is option (D).

3 0
3 years ago
The animal kingdom is divided into two small groups name these two smaller groups and give examples of each include the characte
Elina [12.6K]

Answer:

vertebrate and invertebrates

Explanation:

All Vertebrates  have a skeletal structure with a spinal backbone and well developed internal skeleton made up of bones and cartilage while Invertebrates have no backbone.

Vertebrate include fish, amphibians, reptiles, birds and mammals. Invertebrates include worms and jellyfish, spiders and crabs.

<u>common characteristics among vertebrates and invertebrates </u>

○ multi-cellular.

○ heterotrophs (cannot make their own food) and must get their energy by consuming plants or other animals.

○ obtain food and oxygen for energy,

5 0
3 years ago
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