Ask question

Ask question

All categories

2 years ago

15

What is the least informative property of a mineral

1 answer:

Scrat [10]2 years ago

6 0

All of them are important to identify a mineral and get information from it, but I'd say the least important is cleavage.

You might be interested in

С каким из приведенных веществ взаимодействует соляная кислота

Alexus [3.1K]

Answer:

Соляная кислота оказывает разъедающее действие на глаза, кожу и слизистые оболочки. ... Обычными химическими веществами, добавляемыми к этим жидкостям, являются полимеры для повышения вязкости ... Производятся в промышленных масштабах путем взаимодействия хлорида натрия и серной кислоты; ... объем разбавляется 85% водой в исходном растворе перед закачкой в пласт.

Explanation:

6 0

3 years ago

(25 hours ) /1 ( minutes) = ( hour )

Finger [1]

60 minutes
thanks again I appreciate you I'm sorry but you ok if I invite to you in my prayers for you and Mani

3 0

3 years ago

What happens when magma cools either underground or on the surface and hardens?

exis [7]

I think it turns into an igneous rock

6 0

3 years ago

Read 2 more answers

In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of t

Radda [10]

<u>Answer:</u> The empirical formula for the given compound is $CuCl_3$

<u>Explanation:</u>

We are given:

Mass of copper chloride in 1 L or 1000 mL of solution = 42.62 grams

<u>Taking Trial A:</u>

Volume of solution = 49.6 mL

Applying unitary method:

In 1000 mL of solution, the mass of copper chloride present is 42.62 grams

So, in 49.6 mL of solution, the mass of copper chloride will be = $\frac{42.62}{1000}\times 49.6=2.114g$

We are given:

Mass of filter paper = 0.908 g

Mass of filter paper + copper = 1.694 g

Mass of copper = [1.694 - 0.908] g = 0.786 g

Mass of chlorine in the sample = [2.114 - 0.786]g = 1.328 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Copper = $\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{0.786g}{63.5g/mole}=0.0124moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{1.328g}{35.5g/mole}=0.0374moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0124 moles.

For Copper = $\frac{0.0124}{0.0124}=1$

For Chlorine = $\frac{0.0374}{0.0124}=3.02\approx 3$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Cu : Cl = 1 : 3

Hence, the empirical formula for the given compound is $CuCl_3$

3 0

3 years ago

Given the reaction:

Nadusha1986 [10]

Your answer will be B

8 0

3 years ago

Read 2 more answers