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kvv77 [185]
3 years ago
12

Which best describes the importance of the microscope to the development of cell theory? Microscopes color different parts of ce

lls so they can be seen with the naked eye. Cells are too small to see with the naked eye, and a microscope magnifies them. The lenses in a microscope reduce the image of an organism to the size of a cell. Microscopes trap “animalcules” from pond water so they can be studied.
Chemistry
2 answers:
aliya0001 [1]3 years ago
8 0
Cells are too small to see with the naked eye. 

It's pretty straight forward, use the cross-out method. 

1) Microscopes MAGNIFY images, they don't color the cells. In fact, scientists have to use these chemicals to "stain" or color the cells to see them more easily through microscopes. 

2) If the lenses of a microscope reduced the image of an organism to the size of a cell, you'd be seeing a very tiny human through your microscope, instead of actual cells. 

3) Microscopes don't "trap" anything. In fact, scientists use plates or slides under microscopes to contain what they're studying. 

Elza [17]3 years ago
5 0

Answer:The correct answer is B hope this helps

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The potential energy diagram for a reaction starts at 180 kJ and ends at 300 kJ. What type of reaction does the diagram best rep
aleksklad [387]

Answer: Endothermic reaction

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and \Delta H for the reaction comes out to be negative.

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and \Delta H for the reaction comes out to be positive.

As the energy of reactants is 180 kJ and that of products is 300 kJ, the energy of products is greater than that of reactants, which means  the energy has been absorbed and reaction is endothermic.

8 0
4 years ago
How many liters of a 1.5 M solution can you make if you have .50 mol of KCl?
vaieri [72.5K]

Answer:

The answer to your question is 0.33 liters

Explanation:

Data

Volume = ?

Molarity = 1.5 M

number of moles = 0.5

Formula

Molarity = \frac{number of moles }{volume}

Solve for V

Volume = \frac{number of moles}{molarity}

Substitution

Volume = \frac{0.5}{1.5}

Simplification and result

       Volume = 0.33 l

7 0
3 years ago
Un deportista de 72 kg trepa por una cuerda hasta una altura de 12m. Calcula el incremento de energía potencial gravitatoria que
Oxana [17]

A 72 kg athlete climbs a rope to a height of 12m. Calculate the increase in gravitational potential energy it has experienced.

Answer:

8467.2J

Explanation:

Given parameters:

Mass of the athlete = 72kg

Height of the climb  = 12m

Unknown:

Increase in gravitational potential energy it has experienced = ?

Solution:

Gravitational potential energy is the energy due to the position of a body. It is mathematically expressed as;

  Gravitational potential energy  = m x g x h

m is the mass

g is the acceleration due to gravity  = 9.8m/s²

h is the height

   Insert the parameters and solve;

  Gravitational potential energy  = 72 x 9.8 x 12

  GPE = 8467.2J

4 0
3 years ago
Which reaction is single replacement?
Tatiana [17]
<span>Among the given choices, the third option is the only one which illustrates single replacement. (3)H2SO4 + Mg --> H2 + MgSO4 A single replacement is also termed as single-displacement reaction, a reaction by which an element in a compound, displaces another element. It can be illustrated this way: X + Y-Z → X-Z + Y</span>
5 0
3 years ago
Read 2 more answers
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

3 0
3 years ago
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