<span>The
kingdom, protista’s characteristics are that the organism (not a plant,
animal or fungus) are:
unicellular however some are multicellular like algae, are heterotrophic or
autotrophic, others lives in water while some live in moist areas or human body,
have a nucleus, cellular respiration is primarily aerobic, some are pathogenic
(e.g. causing Malaria) and reproduction is mitosis or meiosis. This kingdom
includes: Sacordinians – pseudopods (e.g. Amoeba, Foraminiferans<span>.)</span>, Zooflagellates – flagellates
(e.g. Trypanosoma gambiense),
Ciliaphorans – ciliates (e.g. paramecium) and Sporozoans (e.g. Plasmodium).</span>
An aqueous solution of an arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt.
<h3>What is a Salt?</h3>
This is a compound which is formed as a result of a neutralization reaction between acid and base.
Arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt due to increased concentration of H+ and OH- respectively.
Read more about Salt here brainly.com/question/13818836
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<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
Answer: 502 Joules
Explanation:
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 40.0 mL
Putting values in above equation, we get:
When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
q = heat absorbed by water
= mass of water = 40.0 g
= final temperature of water = 20.0°C
= initial temperature of water = 17.0°C
= specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
![q=40.0\times 4.186\times (20.0-17.0)]](https://tex.z-dn.net/?f=q%3D40.0%5Ctimes%204.186%5Ctimes%20%2820.0-17.0%29%5D)
Hence, the joules of heat were re-leased by the lead is 502
