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3 years ago

Two grams of potassium chloride are completely dissolved in a sample of water in a beaker.this solution is classified as

1 answer:

8 0

It is classified as a homogeneous mixture because you cannot see any of the parts of the solution because they've been mixed perfectly.

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Suppose you throw a cube-shaped object and a spherical-shaped object of equal volume in the same volume of water. The ball sinks

brilliants [131]

I am pretty sure the answer is B but correct me if I'm wrong hope this helps.

6 0

3 years ago

Which of the following is a mixture steel water oxygen or gold

MAXImum [283]

Steel is an alloy and consists of more than one element (mainly iron and carbon), therefore steel is a mixture.

7 0

3 years ago

Calculate the energy required to melt 21 g of ice at 0 oC.

vladimir1956 [14]

We are given:

Mass of ice = 21 grams

The ice is already at 0°c, the temperature at which it melts to form water

Molar heat of fusion of Ice = 6.02 kJ/mol

Finding the energy required:

Number of moles of Ice:

Molar mass of water = 18 g/mol

Number of moles = given mass/ molar mass

Number of moles = 21 / 18 = 7/6 moles

Energy required to melt the given amount of ice:

Energy = number of moles * molar heat of fusion

Energy = (7/6) * (6.02)

Energy = 7.02 kJ OR 7020 joules

7 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$

(b) Since, the rate equation is as follows.

Rate = $K [OCl^{-}][l^{-}]$

Let us assume that ( $[OCl^{-}] = [l^{-}]$ )

Putting the given values into the above equation as follows.

$1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2$

$1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})$

K = $\frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}$

= $60.4 M^{-1}sec^{-1}$

Hence, the value of rate constant for the given reaction is $60.4 M^{-1}sec^{-1}$ .

(c) Now, we will calculate the rate as follows.

Rate = $K [OCl^{-}][l^{-}]$

= $60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})$

= $6.52 \times 10^{5}$

Therefore, rate when $[OCl^{-}] = 1.8 \times 10^{3}$ M and $[I^{-}]= 6.0 \times 10^{4}$ M is $6.52 \times 10^{5}$ .

8 0

3 years ago

__SIO2+__HF--> __ H2SIF6+ __H2O balanced equation

Dima020 [189]

1,6,1,2 that should be the answer

6 0

3 years ago