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3 years ago

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An irregular object with a mass of 18.00 kg displaces 2.50 L of water when placed in a large overflow container. Calculate the d

ensity of the object.

1 answer:

kolezko [41]3 years ago

7 0

Answer:

7.20 kg/L

Explanation:

Density is mass (kg) / volume (L).

18.00kg/2.50L = 7.20 kg/L

If you need it in g/mL this is the equation:

$\frac{7.20kg}{L} *\frac{1000g}{1kg} *\frac{1L}{1000mL} = 7.20g/mL$

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3 years ago

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The cis and trans geometric isomers of a compound have different chemical and physical properties. Draw the structure of the geo

frutty [35]

<u>Answer:</u> The structure of the geometrical isomers are attached below.

<u>Explanation:</u>

Cis- and Trans- isomers are the geometrical isomers which have same chemical formula but different structural formula

According to CIP rule, the groups on the doubly bonded carbon atoms are given priorities based on the the atomic masses of first connected atom.

If the highest priority groups are on the same side, it is known as cis-form and if the highest priority groups are on opposite side, it is known as trans-form.

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The structure of the geometrical isomers are attached below.

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3 years ago

How would the boiling point of water be affected if you doubled the amount of water?

Flauer [41]

Answer:

The boiling point wouldn´t be effected. The amount of time it would take to get it to that point would increase

Explanation:

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3 years ago

I WILL GIVE 15 POINTS PLS I NEED THIS BEFORE THE CLASS FINISHES IN 30 MINN

bulgar [2K]

Answer:

likely be the same

Explanation:

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3 years ago

How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

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3 years ago