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Anuta_ua [19.1K]

3 years ago

12

Copper sulfate is made of one copper (Cu) atom, one sulfur (S) atom, and four oxygen (O) atoms. Write the chemical formula corre

ctly.

1 answer:

Rzqust [24]3 years ago

4 0

The chemical formula for copper sulfate is CuSO4

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A chemical equation that does not indicate relative amounts of reactants and products is called

Zielflug [23.3K]

Answer:

The correct answer is skeleton equation.

Explanation:

In chemistry, the skeletal formula of a compound is an abbreviated representation of its molecular structure. Skeleton formulas are used because they clearly show complicated structures, they are fast and simple to draw.

All atoms that are not carbon or hydrogen are represented by their chemical symbol. The relative amounts of reagents and products are not indicated.

Have a nice day!

6 0

3 years ago

Differentiate between crystalline and amorphous solid.

Anit [1.1K]

<h2>DIFFERENCE BETWEEN CRYSTALLINE AND AMORPHOUS SOLIDS :</h2><h2><em><u> Amorphous solids do not have definite melting points but melt over a wide range of temperature because of the irregular shape. Crystalline solids, on the other hand, have a sharp melting point.</u></em></h2>

4 0

3 years ago

Read 2 more answers

The rate of a hypothetical reaction involving L and M is found to double when the concentration of L is doubled and to increase

Semmy [17]

Answer: $Rate=k[L]^1[M]^2$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$L+M\rightarrow products$

$Rate=k[L]^x[M]^y$ (1)

k= rate constant

x = order with respect to L

y = order with respect to M

n =( x+y)= Total order

a) If [L] is doubled, the reaction rate will increase by a factor of 2:

$2\times Rate=k[2L]^x[M]^y$ (2)

b) If [M] is doubled, the reaction rate will increase by a factor of 4:

$4\times Rate=k[L]^x[2M]^y$ (3)

Dividing 2 by 1:

$\frac{2\times Rate}{Rate}=\frac{k[2L]^x[M]^y}{k[L]^x[M]^y}$

$2=2^x$

$x=1$

Dividing 3 by 1

$\frac{4\times Rate}{Rate}=\frac{k[L]^x[2M]^y}{k[L]^x[M]^y}$

$4=2^y$

$2^2=2^y$

$y=2$

Thus rate law is: $k[L]^1[M]^2$

4 0

4 years ago

How long does it take for a purple stem plant to grow

Gala2k [10]

I couldn't really find anything about the growth time but it does say that it could remain viable in soil for up to 40 years

4 0

4 years ago

The Mystery Salt

ryzh [129]

Question 1 :

To determine whether the salt is KCl or KNO3, one should look for the difference between them in terms of their physicochemical properties, such as their solubility.

Since we have the solubilité of KCl and KNO3, we can use the property of solubility to determine if the mystery salt is KCl or KNO3.

Question 2:

We will try to reproduce the conditions to determine the solubility of the salt at 37°C.

We will put into the beaker 100ml of water (equivalent to 100g) and dissolve a defined quantity of the salt (the number should be between the solubility of the KCl (37g) and KNO3 (30g) so between 30g and 37g).

Let's dissolve for example 32g of the salt, then, heat with the hotplate until the temperature of the beaker content will be 35 ° C (use the thermometer to determine the exact temperature).

Why?

This manipulation aims to determine the solubility of our mystery salt to know if it is KNO3 or KCl. In our conditions, we will obtain two different possibilities depending on if the salt is KCl or KNO3, this justifies why we took a quantity between 30g and 37g of salt.

If it is KNO3 (solubility of 30g/ml) we will observe a precipitation in the beaker because we exceed its solubility.

If it is KCl (37g/100) we will not observe a precipitate since we did not attempt the solubility of KCl

Question 3:

Finally to determine the composition of salt: we know that the solubility of KCL is 37g / 100ml (that is to say if we dissolve a higher mass (38g for example), we will observe a precipitation of salt) and that the solubility of KNO3 is 30g / 100ml (that is to say if we dissolve an upper mass (32g for example), we will observe a precipitation of salt)

In our experiment, 32g of salts were dissolved. If it is KCl, we will not observe a precipitate since the minimum concentration to start having a precipitate is not yet reached (37g / 100ml).

If it is KNO3, a precipitate will be observed since the minimum concentration to start having a precipitate is not yet reached (30g / 100ml).

6 0

3 years ago