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fgiga [73]
3 years ago
13

A mutation that prevents a maple tree frome efficiently taking gases from the air would most directly affect which of the follow

ing processes? a. reproduction b. photosynthesis c. water uptake d. DNA replication​
Chemistry
1 answer:
harkovskaia [24]3 years ago
7 0

Answer:

b. photosynthesis

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What are the four season of the northern hemisphere .
Ksju [112]

Answer:

spring, summer, fall, and winter

5 0
3 years ago
Read 2 more answers
Prove that PV = nRT.​
qaws [65]

Find your answer in the explanation below.

Explanation:

PV = nRT is called the ideal gas equation and its a combination of 3 laws; Charles' law, Boyle's law and Avogadro's law.

According to Boyle's law, at constant temperature, the volume of a gas is inversely proportional to the pressure. i.e V = 1/P

From, Charles' law, we have that volume is directly proportional to the absolute temperature of the gas at constant pressure. i.e V = T

Avogadro's law finally states that equal volume of all gases at the same temperature and pressure contain the same number of molecules. i.e V = n

Combining the 3 Laws together i.e equating volume in all 3 laws, we have

V = nT/P,

V = constant nT/P

(constant = general gas constant = R)

V = RnT/P

by bringing P to the LHS, we have,

PV = nRT.

Q.E.D

6 0
3 years ago
According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams
gogolik [260]

Answer:

m_{HClO_3}=12.7gHClO_3

Explanation:

Hello,

Considering the reaction:

3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3

The molar masses of chlorine and chloric acid are:

M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3

Best regards.

4 0
4 years ago
Match each land resource to its use.
finlep [7]

Answer:

Explanation:

We are to match each land resource to what they are being used for.

Clay             →→→  used to make pottery and tiles

iron ore       →→→  used to make steel

Salt              →→→   used as a flavoring in food

aggregate   →→→   used in construction

graphite       →→→ used to make batteries

Clay is a kind of soil particle that forms as a result of weathering processes. Examples include; pottery clays, glacial clays, and deep-sea clays e.t.c. The presence of one or more clay minerals, as well as variable quantities of organic and detrital components, characterizes all of them. Clay is usually sticky and moist when wet, but hard when dry. They are used in the making of tiles and potteries.

Iron ore:  The iron ore deposits are found in the Earth's crust's sedimentary rocks. They're made up of iron and oxygen that mix during the chemical process in marine and freshwater. iron ores are used to produce almost every iron and steel product that we use today.

Aggregate: are utilized in construction activities. It is a material used to mix cement, gypsum, bitumen, or lime to produce concrete in the construction industry.

Graphite: Graphite is a mineral that occurs in both igneous and metamorphic rocks. It is generally generated on the earth's surface when carbon is exposed to high temperatures and pressures. It is mainly used in the production of batteries and electrodes,

3 0
3 years ago
A 52 gram sample of an unknown metal requires 714 Joules of energy to heat it from
ratelena [41]

Answer:  Approximately 0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}

===================================================

Work Shown:

We have the following variables

  • Q = 714 joules = heat required
  • m = 52 grams = mass
  • c = specific heat = unknown
  • \Delta t = 82-30.5 = 51.5 = change in temperature

note: the symbol \Delta is the uppercase Greek letter delta. It represents the difference or change in a value.

Apply those values into the formula below. Solve for c.

Q = m*c*\Delta t\\\\714 = 52*c*51.5\\\\714 = 52*51.5*c\\\\714 = 2678*c\\\\2678*c = 714\\\\c = \frac{714}{2678}\\\\c \approx 0.26661687826737\\\\c \approx 0.267\\\\

The specific heat of the unknown metal is roughly 0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}

3 0
3 years ago
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