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lora16 [44]
3 years ago
5

A chemist prepares a solution of copper(II) sulfate by measuring out 17.2 of copper(II) sulfate into a volumetric flask and fill

ing the flask to the mark with water. Calculate the concentration in of the chemist's copper(II) sulfate solution.
Chemistry
2 answers:
diamong [38]3 years ago
7 0

Answer:

Concentration of the chemist's copper(II) sulfate solution is 0.43 mol/L

<em>Note: The volume of the flask is assumed  to be 250 mL</em>

Explanation:

Since the volume of the volumetric flask is not given, we assume the volume its volume to be 250 mL.

Mass of copper (ii) sulfate = 17.2 g; molar mass of copper (ii) sulfate = 160 g/mol

Concentration in mol/L = number of moles/volume in litres

Number of moles of copper (ii) sulfate = 17.2 g / 160 g/mol = 0.1075 moles

Volume of flask in litres = 250 mL/1000 mL * 1 L = 0.250 L

Concentration = 0.1075 moles / 0.250 L = 0.43 mol/L

Therefore, concentration of the chemist's copper(II) sulfate solution is 0.43 mol/L

pshichka [43]3 years ago
5 0

Answer:

0.5000

Explanation:

We can measure the concentration of the copper sulfate solution as 1.000 mol/L. Since the tank contains one half liter, the solution contains 0.5000 moles of copper sulfate.

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K + Li2O what is the product
shutvik [7]

Answer:

2K + Li2O → 2Li + K2O

3 0
3 years ago
Calcium chloride contains calcium and chloride ions. Write the ground-state electron configuration for the calcium ion.
Drupady [299]

Answer:

1s2 2s2 2p6 3s2 3p6

Explanation:

Calcium is the 20th element. It has 20 electrons in it's neutral state. The electronic configuration is given as;

1s2 2s2 2p6 3s2 3p6 4s2

However in the compound; calcium chloride, it does not exist in it's neutral state.

CaCl2 --> Ca2+   +   Cl-

In Ca2+ state, it has lost two electrons, The total number of electrons becomes; 20 - 2 = 18 electrons.

The electronic configuration is given as;

1s2 2s2 2p6 3s2 3p6

8 0
3 years ago
zn (s) 2 hcl (aq) -----&gt; zncl2 (aq) h2(g) if 520 ml of h2 is collected over water at 28oc and the atmospheric pressure is 1.0
nadya68 [22]

The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.

Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.

We recognise that

h * 2 = PT - P * h * 20 = 1atm - 0.037atm

= 0.963 atm

1 * h * 2 = Ph * 2V / R * T

= 0.963 atm x 0.520 L / 0.0821 L atm/

molK * 301

= 0.02 mol h2

= 0.02molZn

So 0.02 mol Zn x 65.39 g/mol

= 1.33 g Zn

Read more about zinc;

brainly.com/question/28880469

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3 0
1 year ago
In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. 4NH3 + 5O2 &gt; 4NO + 6H2O a. Which reactant is the limi
cluponka [151]

Answer:

2.61 g of NO will be formed

The limiting reagent is the O₂

Explanation:

The reaction is:

4NH₃  +  5O₂  →  4NO  +  6H₂O

We convert the mass of the reactants to moles:

3.25g / 17 g/mol = 0.191 moles of NH₃

3.50g / 32 g/mol =0.109 moles of O₂

Let's determine the limiting reactant by stoichiometry:

4 moles of ammonia react with 5 moles of oxygen

Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent

Ratio with NO is 5:4

5 moles of oxygen produce 4 moles of NO

0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO

We convert the moles to mass, to get the answer

0.0872 mol . 30g / 1 mol = 2.61 g

7 0
3 years ago
Give an example of how natural selection could benefit a species. Justify your response in two or more complete sentences.
Sunny_sXe [5.5K]

Answer and Explanation:

Natural selection can benefit a species in many ways. One way natural selection benefits a species is by helping a species adapt to constant changing environments and biomes. natural selection can also benefit a species by dying of the weaker links of a species and adapting the mutated species

7 0
3 years ago
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