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Studentka2010 [4]
4 years ago
12

What do you use to test ph in swimming pool

Chemistry
1 answer:
Alona [7]4 years ago
7 0

Answer:

to make sure the cholerie is not to high

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What is the importance of dissolved co2 and o2 in water
hodyreva [135]

Answer:

The oxygen O2, and the carbon dioxide (CO2) dissolved in water are very important to keep live in aquatic bodies.

Explanation:

Oxygen (O2) is an element found in nature usually as a gas in the air that we breathe. Oxygen is very important for aerobic organisms, including us, the humans. It can dissolve in water, and it is vital for the health of aquatic habitats. If we have low quantity of oxygen dissolve in water means that could be a pollution problem. Actually the oxygen level is an indicator to determine the water quality. In order to survive the aquatic biota (animals and plants) need certain quantity of oxygen. The oxygen dissolved in the water comes from the air, because of the wind, rain, waves, currents, and from aquatic plants like algae thanks to a process call photosynthesis.

Carbon Dioxide (CO2) is a chemical compound. This gas also is found in the air, and can be dissolved in water easier than oxygen. Aquatic plants needs CO2, to produce their own food during the photosynthesis process. It´s very important to have enough CO2, for the growth of aquatic plants. But if we have to much CO2 the fish will not have enough oxygen and could suffocate and die. Aquatic animals through respiration process produce CO2, it also can be obtained from fermentation, and from the burning of any substance that has carbon C.

If we want to preserve live in the aquatic bodies we must keep the balance between oxygen and carbon dioxide.

4 0
3 years ago
To begin the experiment, 1.11g of methane is burned in a bomb container containing 1000grams of water. The initial temperature i
PilotLPTM [1.2K]
The formula of heat added is presented as
Q = m * c * change in (T)

where 
Q - the heat added
m - mass
c - specific heat
<span>Change in temperature</span> - (Tf - To)

Basically, in order for us to know the change in heat, the formula is
Change in temperature = T(final) - T(original)
Change in temperature<span> = 36.65 - 24.85
</span>Change in temperature<span> = 11.8 degrees Celcius
</span>
So the change in temperature is 11.8 degrees 
8 0
4 years ago
Read 2 more answers
Please help me asap​
Masja [62]

Answer: With the first quarter moon, the Moon is "waxing", which means getting larger. with the third quarter moon, the Moon is "waning", which means getting smaller.

Explanation: In each case, the Moon is half lit ---- first quarter moon is when it is increasing and the third quarter moon is decreasing

Hope this helped! :)

7 0
3 years ago
Calculate the molarity of 29.9 g of MgS in 911 mL of solution??
LUCKY_DIMON [66]

0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Given data:

Volume of solution = 911 mL=0.911L

Given the mass of  MgS = 29.9 g

Molar mass of MgS = 56.38 g/mol

Moles of MgS = 0.5303299042

Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}

Molality = \frac{0.5303299042}{0.911L}

Molality = 0.5821403998=0.58

Hence,  0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Learn more about Molarity here:

brainly.com/question/2817451

#SPJ1

3 0
2 years ago
Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.
natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

pH=4.76+log(\frac{0.95}{0.92})

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

pH=4.76+log(\frac{0.99}{0.88})

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

pH=4.76+log(\frac{0.85}{1.02})

pH = 4.76 - 0.079

pH = 4.681

5 0
4 years ago
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