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Sergeeva-Olga [200]

3 years ago

15

Cubic zirconia is a substance produced in a laboratory. It looks nearly identical to a diamond. Why is it not a mineral?

2 answers:

galina1969 [7]3 years ago

7 0

The last one, not natuslly occuring

Lana71 [14]3 years ago

5 0

Answer:

it is not naturally occuring

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You can detect salt in water without tasting by measuring the density of the water. Place a glass of spring water and a glass of the suspected salt water on a balance scale and the heavier one contains salt. Other ways to test for salt in water is to put a drop of water on the end of a nail and place in a gas flame. If the water contains salt, the flame will turn a yellow/orange color.

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3 years ago

What is the name of Bel on the periodic table

rewona [7]

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Nobelium or Beryllium

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2 years ago

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liraira [26]

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3 years ago

Question 116 g of sunflower oil is

Luba_88 [7]

<h2>Answer: 125.41 mL</h2>

Explanation:

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= 116 g ÷ 0.925 g/mL

= 125.41 mL

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7 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago