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Vikentia [17]
2 years ago
13

A root is an example of a/an- A: tissue B: cell type C:organ D:organism

Chemistry
2 answers:
konstantin123 [22]2 years ago
7 0

Answer: D an organism

Explanation:

sveta [45]2 years ago
5 0
It would be classified as an organ

the answer is C.

have a good day mate
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If you set up an experiment with two different independent variables, then the results would be_____.
MariettaO [177]

Answer: A) Inconclusive; you would not know which of the two variables caused the change.

Explanation:

When you set up an experiment, you must make sure that you control the variables such that only one independent variable changes at a time, while all the remainder conditions (the other independent variables) are controlled (fixed).

By observing (measuring) the dependent variable, while only one independent variable changes you can understandhow such independent variable explains (determines) the dependent variable, leading to a conclusion.

Conversely, if two or more independent variables change at a time, then there is no way that you can tell how the output (dependent variable) is related with one or other of the changes of the indipendent variables. You wolud not be able to discriminate (distinguish) the effect of one or other variable, making the experiment inconclusive

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4 0
3 years ago
Why cant magnesium ions be detected in a flam test?
Dmitry [639]
When magnesium ion doesn't give any characteristics colour with the flame test as electronic transisitons do not give out visible light.
6 0
3 years ago
Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O
Bess [88]
40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O
5 0
3 years ago
How and why are Endothermic and Exothermic reactions chemical changes?
Olin [163]

Answer:

Chemical reactions often involve changes in energy due to the breaking and formation of bonds. Reactions in which energy is released are exothermic reactions, while those that take in heat energy are endothermic.

Explanation:

8 0
3 years ago
Caffeine, a stimulant found in coffee and soda, hasthe mass percent composition: C. 49.48%, H, 5.19%. N. 28.85% 0. 16.48% The mo
borishaifa [10]

We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

C_4H_5O_1N_2

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

C_8H_{10}O_2N_4

8 0
1 year ago
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