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Serggg [28]
4 years ago
6

A gas has a volume of 27.5 L at 302 K and 1.40 atm. How many moles are in the sample of gas?

Chemistry
1 answer:
Temka [501]4 years ago
7 0

it may be C. did you get the right answer so you could study it?

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Plants need sunlight to make food during the process of:
Andrej [43]

the answer is  photosynthesis

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Which is the correctly balanced chemical equation for the reaction of KOH and H2SO4? Upper K upper O upper H right arrow upper H
sertanlavr [38]

Answer :

The balanced chemical reaction will be,

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

If the amount of atoms of each type on the left and right sides of a reaction differs then to balance the equation by adding coefficient in the front of the elements or molecule or compound in the chemical equation.

The coefficient tell us about that how many molecules or atoms present in the chemical equation.

When sulfuric acid react with potassium hydroxide then it react to give potassium sulfate and water as a product. This reaction is known as acid-base reaction.

The balanced chemical reaction will be,

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

4 0
4 years ago
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The air we breathe is not considered by chemists to be an element. why not
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Because Air itself is not an element. it is made up by different elements such as Oxygen and nitrogen. since its made up of a mix of different elements, its rather a homogeneous mixture :)
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3 years ago
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[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con
telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = 284 \ grams

solid soil volume =205 \ cc

saturated mass soil = 361 \ g

The weight of the soil after drainage is =295 \ g

Water weight for soil saturation = (361-284) = 77 \ g

Water volume required for soil saturation =\frac{77}{1} = 77 \ cc

Sample volume of water: = \frac{\text{water density}}{\text{water density input}}

= 361- 295 \\\\ = 66 \ cc

Soil water retained volume = (draining field weight - dry soil weight)

                                             = 295 - 284 \\\\ = 11 \ cc.

\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}

                    = \frac{77}{(205 + 77)}  \\\\= \frac{77}{(282)}  \\\\ = 27.30 \%

(Its saturated water volume is equal to the volume of voids)

\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}

                              = \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23

\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}

                            = \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04

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3 years ago
an ecologist who is studying the mineral requirements and daily water intake of a plant species is studying
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They are studying the planet's ecosystem.
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