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3 years ago

Plant use sunlight as energy to convert carbon dioxide and water into glucose and oxygen. Which best describes the reaction?

Chemistry

2 answers:

Illusion [34]3 years ago

8 0

Answer:

Its the process of photosynthesis

motikmotik3 years ago

3 0

Answer:

photosynthesis(definitely the answer) or photolysis(different, slightly, but kinda the same thing)

Explanation:

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bija089 [108]

Colorful flowers and blueberrie looking ones and the trees

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3 years ago

Read 2 more answers

4/5(2x 5)−4=1

uysha [10]

5/8=0.625 4/5=.8

5/8 because 5/8=0.625 and because its 2x you multiply 0.625 by 2=1.25 then you do 1.25+5=6.25 4/5=.8 6.25 times .8=5 5-4=1

X=5/8 or 0.625

5 0

3 years ago

A renowned laboratory reports quadruple-point coordinates of 10.2 Mbar and 24.1°C for four-phase equilibrium of allotropic solid

Inessa [10]

Answer:

The claim is untrue.

Explanation:

Given the information from the question. We need to evaluate the claim.

According to the phase rule we have F= C-P+2

In this particular situation, the forms are completely allotropic .In order words, they are conjured through the same chemical composition. Thus constituents C= 1, P=4 for four phases and the number variables is 2. As a result, F= C-P+2 =1-4+2= -1. Therefore, the claim is untrue.

8 0

3 years ago

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0

3 years ago

[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)

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

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

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

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

6 0

3 years ago