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2 years ago

9

Many protists are single-celled organisms. Which of the following characteristics is also common to organisms in Kingdom Protist

a?
A.
thrive in environments without oxygen

B.
thrive in environments with little sunlight

C.
thrive in wet environments

D.
thrive in environments without water

2 answers:

Alenkinab [10]2 years ago

8 0

THE ANSWER IS C I JUST TOOK THE TEST. THE ANSWER IS ::::

thrive in wet environments

HACTEHA [7]2 years ago

6 0

Protists belong to the group eukaryotes (having their DNA enclosed inside the nucleus). They are not plants, nimals or fungi but they act like one. They can be in general subgroups such as unicellular algae, protozoa and molds. They thrive in environments with little sunlight. The answer is letter B.

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A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027

irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

$C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\$

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

x _________ 25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

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3 years ago

Explain why a total lunar eclipse appears red.<br> I

DochEvi [55]

Answer:

During a total lunar eclipse, the Earth lies directly between the sun and the moon, causing the Earth to cast its shadow on the moon.

Explanation:

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3 years ago

"Compounds A and B react to give a single product, C. Write the rate law for each of the following cases and determine the units

olasank [31]

Answer:

Part a: <em>Units of k is </em> $M^{-2}s^{-1}$ <em> where reaction is first order in A and second order in B</em>

Part b: <em>Units of k is </em> $M^{-1}s^{-1}$ <em> where reaction is first order in A and second order overall.</em>

Part c: <em>Units of k is </em> $M^{-1}s^{-1}$ <em> where reaction is independent of the concentration of A and second order overall.</em>

Part d: <em>Units of k is </em> $M^{-3}s^{-1}$ <em> where reaction reaction is second order in both A and B.</em>

Explanation:

As the reaction is given as

$A+B \rightarrow C$

where as the rate is given as

$r=k[A]^x[B]^y$

where x is the order wrt A and y is the order wrt B.

Part a:

x=1 and y=2 now the reaction rate equation is given as

$r=k[A]^1[B]^2$

Now the units are given as

$r=k[A]^1[B]^2\\M/s =k[M]^1[M]^2\\M/s =k[M]^{1+2}\\M/s =k[M]^{3}\\M^{1-3}/s =k\\M^{-2}s^{-1} =k$

The units of k is $M^{-2}s^{-1}$

Part b:

x=1 and o=2

x+y=o

1+y=2

y=2-1

y=1

Now the reaction rate equation is given as

$r=k[A]^1[B]^1$

Now the units are given as

$r=k[A]^1[B]^1\\M/s =k[M]^1[M]^1\\M/s =k[M]^{1+1}\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part c:

x=0 and o=2

x+y=o

0+y=2

y=2

y=2

Now the reaction rate equation is given as

$r=k[A]^0[B]^2$

Now the units are given as

$r=k[B]^2\\M/s =k[M]^2\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part d:

x=2 and y=2

Now the reaction rate equation is given as

$r=k[A]^2[B]^2$

Now the units are given as

$r=k[A]^2[B]^2\\M/s =k[M]^2[M]^2\\M/s =k[M]^{2+2}\\M/s =k[M]^{4}\\M^{1-4}/s =k\\M^{-3}s^{-1} =k$

The units of k is $M^{-3}s^{-1}$

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3 years ago

Will give brainliest! if 32.0 g of hcl is to be diluted to make a 4.80 m solution, how much water should be added?

nadezda [96]

The required volume of water is 0.18 liters.

<h3>What is molarity?</h3>

Molarity of any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V

Moles of solute will be calculated as:

n = W/M, where

W = given mass of HCl = 32g

M = molar mass of HCl = 36.4g/mol

n = 32 / 36.4 = 0.88 mole

Given molarity of solution = 4.80M

On putting all values in the above equation, we get

V = (0.88) / (36.4) = 0.18 L

Hence required volume of water is 0.18L.

To know more about volume & concentration, visit the below link:

brainly.com/question/26762947

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6 0

2 years ago

A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.

kondaur [170]

The question is incomplete, complete question is:

A fluid occupying $3.2 m^3$ of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = $3.2 m^3$

Density of the fluid = D

$D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3$

Specific volume is the reciprocal of the density :

$V_{specific}=\frac{1}{Density}$

Specific volume of the fluid = $S_v$

$S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg$

2)

Density of the fluid in English Engineering units = D $(lb/ft^3)$

1 kg = 2.20462 lb

1 m = 3.280 ft

$D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3$

Specific volume of the fluid :

$=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb$

3)

Density of the fluid in British Gravitational System units = D $(slug/ft^3)$

1 kg = 0.06852 slug

1 m = 3.280 ft

$D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3$

Specific volume of the fluid :

$=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug$

7 0

3 years ago