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JulijaS [17]
3 years ago
10

Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of Pb and H+ in the bal

anced reaction?
Pb2+(aq) + NH+4(aq) --> Pb(s) + NO-3(aq)
Chemistry
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

kujgedyjmnyged\jmngv\gedkjmyten

Explanation:

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Identify one factor represented in the diagram that is used to classify the ten types of clouds
Ede4ka [16]
Low clouds
Stratus clouds are uniform grayish clouds that often cover the sky. Usually no precipitation falls from stratus clouds, but they may drizzle. When a thick fog “lifts,” the resulting clouds are low stratus. Nimbostratus clouds form a dark gray, “wet” looking cloudy layer associated with continuously falling rain or snow. They often produce light to moderate precipitation.

Middle clouds
Clouds with the prefix “alto” are middle-level clouds that have bases at 6,500 to 23,000 feet up. Altocumulus clouds are made of water droplets and appear as gray, puffy masses, sometimes rolled out in parallel waves or bands. These clouds on a warm, humid summer morning often mean thunderstorms by late afternoon. Altostratus clouds, gray or blue-gray, are made up of ice crystals and water droplets. They usually cover the sky. In thinner areas of them, the sun may be dimly visible as a round disk. Altostratus clouds often form ahead of storms that produce continuous precipitation.

High clouds
Cirrus clouds are thin, wispy clouds blown by high winds into long streamers. They are considered “high clouds,” forming at more than 20,000 feet. They usually move across the sky from west to east and generally mean fair to pleasant weather. Cirrostratus, thin, sheetlike clouds that often cover the sky, are so thin the sun and moon can be seen through them. Cirrocumulus clouds appear as small, rounded white puffs. Small ripples in the cirrocumulus sometimes resemble the scales of a fish, creating what is sometimes called a “mackerel sky.”

Vertical clouds
Cumulus clouds are puffy and can look like floating cotton. The base of each is often flat and may be only 330 feet above ground. The top has rounded towers. When the top resembles a cauliflower head, it is called “cumulus congestus.” These grow upward and if they continue to grow vertically can develop into a giant cumulonimbus, a thunderstorm cloud, with dark bases no more than 1,000 feet above ground and extending to more than 39,000 feet. Tremendous energy is released by condensation of water vapor in a cumulonimbus. Lightning, thunder and violent tornadoes are associated with them.
6 0
3 years ago
What is the linkage that binds one monosaccharide to another?
ra1l [238]

Answer:

hope this helps

Explanation:

glycosidic bond

A covalent bond formed between a carbohydrate molecule and another molecule (in this case, between two monosaccharides) is known as a glycosidic bond (Figure 4). Glycosidic bonds (also called glycosidic linkages) can be of the alpha or the beta type.

4 0
3 years ago
Read 2 more answers
A potential energy diagram is shown.
Vesna [10]

Answer:

25kJ

Explanation:

Given the initial energy to be 30kJ

The energy change from the initial energy to the peak energy = (65-30) kJ

= 35kJ

since the second energy change was a drop in energy it is regarded negative

= (55-65)

= -10kJ

Therefore total energy change

= (35-10)kJ

= 25kJ

5 0
2 years ago
What ion is oxygen likely to form?
Leto [7]
It's most natural state has a charge of -2. So, a negative ion with two more electrons than is normal.
7 0
3 years ago
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
3 years ago
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