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lana66690 [7]
3 years ago
12

Protista are often grouped according to whether they are plant-like or fungus-like. What is another way they can be grouped? Bac

teria-like arcahebacteria-like animal-like virus-like
Chemistry
2 answers:
zalisa [80]3 years ago
7 0

the answer is animal-like, I resiliently took a test on it.

Gelneren [198K]3 years ago
3 0

The right answer is animal-like.

Protists are unicellular living organisms about 0.1 mm in size, such as amoeba, paramecium and euglena.

Protists are a paraphyletic group in the phylogenetic classification. They were, in classical classification, the fourth reign of the domain of eukaryotes but they are currently defined by exclusion, that is to say that they are all eukaryotes belonging neither to mushrooms nor to plants or animals.

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How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (II) nitrate solution r
NNADVOKAT [17]
2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃

m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol

1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw

2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}

m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}

m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}

m(Fe)=55.847*245*0.805/179.857= 61.24 g


3 0
3 years ago
Read 2 more answers
It is difficult to measure the correct atomic radius. Why?
Paul [167]

Answer:

its kinda you just need to simplify

Explanation:

4 0
3 years ago
The mass of an object is 240g and its volume is 60 cm3<br> A. Sink<br> B. Float
Vinil7 [7]

Answer:

A Sink

Explanation:

Anything greater than 1g/cm3 will sink in water

4 0
4 years ago
A sealed piston contains 400 mL of air at standard ambient pressure. The piston is
nydimaria [60]

Answer:

324.24 kPa

Explanation:

Given that;

Initial pressure P1 = 101325 Pa

V1= 400 ml

P2 = ?

V2= 125mL

From Boyle's law;

P1V1 = P2V2

P2 = P1V1/V2

P2= 101325 × 400/125

P2= 324.24 kPa

5 0
3 years ago
Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting
larisa [96]

Answer: Therefore, the concentration of final solution is 4.0\times 10^{-8}M

Explanation:

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 2.3\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

2.3\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=0.28\times 10^{-5}M

Therefore, the concentration of diluted solution is 0.28\times 10^{-5}M

2) On further dilution

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.28\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

0.28\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=0.034\times 10^{-5}M

Therefore, the concentration of diluted solution is 0.034\times 10^{-5}M

3) On further dilution

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.034\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

0.034\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=4.0\times 10^{-8}M

Therefore, the concentration of final solution is 4.0\times 10^{-8}M

5 0
3 years ago
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