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3 years ago

What is an example of a colloid? A. Soda B.Fog C.Jewelry D.Silverware

Chemistry

2 answers:

TEA [102]3 years ago

5 0

Answer:

B. fog

Explanation:

This is because a liquid (water) is dispersed throughout the air (gas)

Olegator [25]3 years ago

4 0

Answer:

the current answer is

silverware

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Which statement is true of the following reaction? H2 + O2 -> 2H+ O2-

Ede4ka [16]

B.it is not balanced for charge or for number of atoms.

6 0

2 years ago

You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne

fenix001 [56]

Answer:

$P_2=1.90atm$

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

$\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}$

Whereas both n and R are cancelled out as they don't change, we obtain:

$\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}$

Thus, by solving for the final pressure, we obtain:

$\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}$

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

$P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm$

Best regards!

7 0

2 years ago

Match each part of the electrochemical cell with its function.

umka21 [38]

Hi!

The correct options would be:

1. Cathode - <em>reduction</em>

The cathode is the negatively charged electrode, and so has an excess of electrons. Cations (positively charged ions) are attracted to the cathode, and gain electrons to acquire a neutral charge. The process in which a gain of electron occurs is called reduction.

2. Anode - <em>oxidation</em>

The opposite occurs at the anode which is positively charged and attracts negatively charged ions, anions. These anions lose their electrons at the anode to acquire a neutral charge, and the process involving loss of electrons is known as oxidation.

3. Salt Bridge - <em>ion transport </em>

Salt bridge is a physical connection between the the anodic and cathodic half cells in an electrochemical cell and is a pathway that facilitates the flow of ions back and forth these half cells. Salt bridge is involved in maintaining a neutral condition in the electrochemical cells, and its absence would result in the accumulation of positive charge in the anodic cell, and negative charge in the cathodic cell.

4. Wire - <em>electron transport </em>

Wires have a universal role of being a pathway for the transport of electrons in circuit. This role is also the same in the wires involved in an electrochemical cells where they are used to transport electrons from the anodic half cell, and this electron transport results in the generation of electricity in the internal circuit of the electrochemical cell.

Hope this helps!

4 0

2 years ago

If 2.0 g of copper(II) chloride react with excess sodium nitrate, what mass of sodium chloride is formed in this double replacem

cluponka [151]

Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CuCl₂: 1 mole
NaNO₃: 2 moles
Cu(NO₃)₂ : 1 mole
NaCl: 2 moles

The molar mass of the compounds is:

CuCl₂: 134.44 g/mole
NaNO₃: 85 g/mole
Cu(NO₃)₂ : 187.54 g/mole
NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
NaNO₃: 2 moles ×85 g/mole= 170 grams
Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
NaCl: 2 moles ×58.45 g/mole= 116.9 grams

<h3>Mass of NaCl formed</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?

$mass of NaCl=\frac{2 grams of CuCl_{2}x116.9 grams of NaCl }{134.44grams of CuCl_{2}}$