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3 years ago

5

All organisms contain special structures that store important information and perform many task to help the organisms survive. T

he picture below shows what some of these structures look like under a microscope.
what are the structures?
A. Cells
B. Air particles
C. Worms
D. Water particles

1 answer:

vova2212 [387]3 years ago

4 0

Answer:

A. cells

Explanation:

i just did the test .

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Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C

Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, $E = \triangle M * 931.5$

$\triangle M = \sum M_{product} - \sum M_{reactant}$

Mass of 1H, $M_{H} = 1.007823$

Mass of 2H, $M_{2H} = 2.0141u$

Mass of 3H, $M_{3H} = 3.016 u$

Mass of Helium, $M_{4He} = 4.002602u$

Mass of Beryllium, $M_{7Be} = 7.01693 u$

Mass of neutron, $M_{n} = 1.008664 u$

a) $2H + 3H \rightarrow 4He + n$

$\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u$

Energy released,

$E = -0.01884 * 931.5\\E = -17.55 Mev$

Energy released = 17.55 MeV

b) $4He + 4He \rightarrow 7Be + n$

$\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} + M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u$

Energy released,

$E = 0.02039 * 931.5\\E = 18.99 Mev$

c) $2H + 2H \rightarrow 3 He$ + n

$\triangle M = (M_{3He} + M_{n} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u$

Energy released,

$E = -0.003536 * 931.5\\E = -3.29 Mev$

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) $2H + 2H \rightarrow 3H + 1H$

$\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u$

$E = -0.00435 * 931.5\\E =-4.075 Mev$

E = 4.075 MeV ( Energy is released)

5 0

3 years ago

a 20.0mL sample of 0.15M hydrochloric acid (HCI) is needed to neutralize a 10.0mL sample of potassium hydroxide (KOH). what is t

Masteriza [31]

Answer: .75 M

Explanation:

8 0

2 years ago

A sample of an unknown gas occupies 5.51 L at 1.31 atm. What pressure would thisgas exert in a 0.520 L container if the temperat

Alex Ar [27]

Answer:

13.9atm

Explanation:

Use Boyle's Law: P1V1 = P2V2

Manipulate the equation: P2 = P1V1/V2

Substitute the values: P2 = (1.31atm)(5.51L)/0.520L

P2 = 13.88096atm

Optionally round to 3 sig figs to get 13.9atm

3 0

3 years ago

Naturally occurring Indium has two isotopes.

trasher [3.6K]

Answer:

Average Molar mass is 114.8u

Explanation:

» From the formula of average Relative Atomic Mass [ RAM ]

${ \rm{RAM = \sum \frac{(isotopic \: mass \times \%abundance)}{100} }} \\$

» We have two isotopes of Indium,

Indium-113, mass » 112.9u, % » 4.28%
Indium-115, mass » 114.9u, % » 95.72%

${ \tt{RAM = \frac{ \blue{(112.9 \times 4.28)} + { \green{(114.9 \times 95.72)}}}{100} }} \\ \\ { \tt{RAM = \frac{11481.44}{100} }} \\ \\ { \underline{ \underline{ \tt{ \: RAM = 114.8\: }}}}$

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2 years ago

How many liters of a normal saline (0.9% NaCl) solution will contain 18 grams of NaCl?

Aleks [24]

838384 liters of pis urin aballs

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2 years ago

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