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omeli [17]
3 years ago
7

If you looked at one of your body's cell under a microscope with 100x magnification, would it be larger, smaller, or the same si

ze as an amoeba?
Chemistry
1 answer:
aev [14]3 years ago
7 0
Peoples cells vary . so theres no definite say
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All physical changes can be identified at the macroscopic level. true or false
Aliun [14]

Answer:

True

Explanation:

The physical changes are reversible in most cases and these changes are not the chemical changes which means that it is only the change in its state not in their nature. Just take the example of water, on cooling it becomes solid and change in color can be seen which is white in solid form and colorless in liquid form. This is also reversible and is a physical change. This means that physical changes can be identified at macroscopic level. Hence the answer is true.

5 0
3 years ago
At 150°C the decomposition of acetaldehyde CH3CHO to methane is a first order reaction. If the
Crank

The decomposition time : 7.69 min ≈ 7.7 min

<h3>Further explanation</h3>

Given

rate constant : 0.029/min

a concentration of  0.050 mol L  to a concentration of 0.040 mol L

Required

the decomposition time

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time

For first-order reaction :

[A]=[Ao]e^(-kt)

or

ln[A]=-kt+ln(A0)

Input the value :

ln(0.040)=-(0.029)t+ln(0.050)

-3.219 = -0.029t -2.996

-0.223 =-0.029t

t=7.69 minutes

4 0
3 years ago
A sample of air is slowly passed through aqueous Sodium hydroxide and then over heater copper. Which gases are removed by this p
bagirrra123 [75]

Carbon dioxide and oxygen are removed from the air.

Explanation:

When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.

First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :

CO₂ + H₂O → H₂CO₃

The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):

H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O

After this by passing the air over heated cooper the oxygen is removed.

2 Cu + O₂ → 2 CuO

Learn more about:

neutralization reaction

brainly.com/question/2632201

#learnwithBrainly

7 0
3 years ago
How US is the most developed country?
8090 [49]
 I believe in what USA was and is able to achieve, and trust the nation in many but every ways. What really made it a developed nation is the things that happened in the last 3 Centuries, some of which are;

#1 CULTURAL TOLERANCE & IMMERSION
#2 TECHNOLOGICAL INVENTIONS & ADVANCEMENT IN THE FIELD OF SCIENCE
#3 FAST FOOD CULTURE
#4 HOLLYWOOD
#5 INTEGRITY & PATRIOTISM
#6 MASS CONSUMERISM
6 0
3 years ago
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
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