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Varvara68 [4.7K]
2 years ago
11

Determine the compound type for the following formulas: C12H22011 Mg(OH)2 H20 Cu3Zn2 Au ​

Chemistry
2 answers:
Scorpion4ik [409]2 years ago
7 0

Answer:

C12H22O11  

✔ covalent

Mg(OH)2    

✔ Ionic

H2O    

✔ covalent

Cu3Zn2    

✔ metallic

Au      

✔ metallic

Explanation:

Marianna [84]2 years ago
5 0

Answer:

Determine the compound type for the following formulas:

C12H22O11

colvent

Mg(OH)2    

inoic

H2O    

colvet

Cu3Zn2  

metallic

Au      

metallic

Explanation:

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What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?
Maru [420]

Answer:

There is 17,114825 g of powdered drink mix needed

Explanation:

<u>Step 1 :</u> Calculate moles

As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L

Since the volume is 100mL, we have to convert the concentration,

⇒0.5 / 1   =  x /0.1    ⇒ 0.5* 0.1  = x = 0.05 M

This means there is 0.05 mol per 100mL

e

<u>Step 2 </u>: calculate mass of the powdered drink

here we use the formula n (mole) = m(mass) / M (Molar mass)

⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.

0.05 mol = mass / 342.2965 g/mol

To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g

There is 17,114825 g of powdered drink mix needed

3 0
2 years ago
Read 2 more answers
7. How many moles of argon are there in 20.0 L, at 25 degrees Celsius and 96.8 kPa?
suter [353]
<h3>Answer:</h3>

              0.8133 mol

<h3>Solution:</h3>

Data Given:

                 Moles  =  n  =  ??

                 Temperature  =  T  =  25 °C + 273.15  =  298.15 K

                  Pressure  =  P  =  96.8 kPa  =  0.955 atm

                  Volume  =  V  =  20.0 L

Formula Used:

Let's assume that the Argon gas is acting as an Ideal gas, then according to Ideal Gas Equation,

                  P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹

Solving Equation for n,

                  n  =  P V / R T

Putting Values,

                  n  =  (0.955 atm × 20.0 L) ÷ (0.082057 atm.L.mol⁻¹.K⁻¹ × 298.15 K)

                 n  =  0.8133 mol

4 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=PCl_5%20%5Crightleftarrows%20PCl_3%20%2B%20Cl_2" id="TexFormula1" title="PCl_5 \rightleftarrow
Butoxors [25]

<u>Answer:</u> The total pressure of the container will be 2.00 atm

<u>Explanation:</u>

We are given:

Initial moles of phosphorus pentachloride = 1.00 atm

For the given chemical reaction:

PCl_5\rightleftharpoons PCl_3+Cl_2

By Stoichiometry of the reaction:

1 mole of PCl_5 produces 1 mole of PCl_3 and 1 mole of chlorine gas

So, 1.00 atm of PCl_5 will also produce 1.00 atm of PCl_3 and 1.00 atm of chlorine gas when the reaction goes to completion.

Total pressure of the container when the reaction goes to completion  = 1.00 + 1.00 = 2.00 atm

Hence, the total pressure of the container will be 2.00 atm

3 0
2 years ago
Which of the following is a correctly written chemical equation that demonstrates the conservation of mass?
Fantom [35]

Answer:

Option D is correct.

H₂O + CO₂      →    H₂CO₃

Explanation:

First of all we will get to know what law of conservation of mass states.

According to this law, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Example:

6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂

there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

Now we will apply this law to given chemical equations:

A) H₂ + O₂   →    H₂O

There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.

B) Mg + HCl   →   H₂ + MgCl₂

In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.

C) KClO₃      →     KCl + O₂

There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.

D)  H₂O + CO₂      →    H₂CO₃

There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.

6 0
2 years ago
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
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