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Mnenie [13.5K]
3 years ago
9

James took two pea plants, placing one in a dark closet and the other on a sunny window sill. Both are located in air conditione

d rooms. He then measured the growth of the pea plants after a week. In this experiment sunlight is a(n) ____.
A. independent variable

B. Constant

C. Dependent variable

D. Control
Chemistry
2 answers:
MariettaO [177]3 years ago
4 0
<span>C. Dependent variable  without the sunlight he could not preform his experiment because if he where to put both in the sunlight the sunlight would then become control but it could never become independent </span>
Umnica [9.8K]3 years ago
3 0

Answer:

independent variable

Explanation:

this is the correct answer

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Find the equation of the line passing through the points (2, 1) and (5, 10).​
Vedmedyk [2.9K]

The equation : y=3x-5

<h3>Further explanation </h3>

Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates

General formula

y-y1 = m(x-x1)

or

y = mx + c

Where

m = straight-line gradient which is the slope of the line

x1, y1 = the Cartesian coordinate that is crossed by the line

c = constant

The formula for a gradient (m) between 2 points in a line

m = Δy / Δx

  • Gradient

\tt \dfrac{10-1}{5-2}=3

  • Equation

\tt y-1=3(x-2)\\\\y-1=3x-6\\\\y=3x-5

3 0
3 years ago
What is the relationship between elements ,atoms ,and compounds
Svetach [21]
Elements are separate particles that contain the properties of only one type of element (pure substance) and an atom represents that element as the smallest non divisible particle that retains the properties of that element. Compounds can be formed by conjoining different atoms together in different ratios and shapes, so a combination of elements.
8 0
3 years ago
Read 2 more answers
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0
3 years ago
What mass of KNO, will dissolve in 100 g of water at 100°C?
Sphinxa [80]

Answer:

About 170-180 grams of potassium nitrate are completely dissolved in 100 g.

Explanation:

Hello!

In this case, according to the reported solubility data for potassium nitrate at different temperatures on the attached picture, it is possible to bear out that about 170-180 grams of potassium nitrate are completely dissolved in 100 g; considering that the solubility is the maximum amount of a solute that can be dissolved in a solvent, in this case water.

Best regards!

3 0
2 years ago
A common fuel additive that is composed of C, H, and O enhanced the performance of gasoline began being phased out in 1999 becau
spayn [35]

Answer:

C₅ H₁₂ O

Explanation:

44 g of CO₂ contains 12 g of C

30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .

18 g of H₂O contains 2 g of hydrogen

14.8 g of H₂0 will contain 1.644 g of  H .

total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O

gram of O = 2.22

moles of C, O, H in the given compound =  8.236 / 12 , 2.22 / 16 , 1.644 / 1

= .6863 , .13875 , 1.644

ratio of their moles = 4.946 : 1 : 11.84

rounding off to digits

ratio = 5 : 1 : 12

empirical formula = C₅ H₁₂ O

6 0
3 years ago
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