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PIT_PIT [208]
3 years ago
11

Diffrent between bryophytes and pteriodphytes

Chemistry
1 answer:
Pie3 years ago
4 0
I would help if i could im sorry im sure someone else will answer soon im honestly trying to help as many people as i can
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What do you expect will occur as air masses move over cold locations? Hot locations? How does this movement of air masses affect
Maksim231197 [3]

Answer:

Explanation:

When air masses will move over cold location then bottom layer of air cools and become more dense. Due to its high density it is trapped near the ground then it flow towards the equator.

When it moves over hot location then bottom layer gets hot and lighter. Then it moves towards poles.

It changes the temperature and humidity of the climate. making hoter region coll and cooler region a bit hot.

3 0
3 years ago
To what volume should you dilute 55 mL of 12 M stock HNO3 solution to obtain a 0.145 HNO3 solution?
velikii [3]

Answer:

4552 mL

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 55 mL

Molarity of stock solution (M₁) = 12 M

Molarity of diluted solution (M₂) = 0.145 M

Volume of diluted solution (V₂) =?

The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

12 × 55 = 0.145 × V₂

660 = 0.145 × V₂

Divide both side by 0.145

V₂ = 660 / 0.145

V₂ ≈ 4552 mL

Thus, the volume of the diluted solution is 4552 mL

7 0
3 years ago
Which reactant (X) is missing from the equation shown?<br> X+ PO4 → H3PO4
insens350 [35]

Answer:

H

Explanation:

b/c the polyatomic PO4¯³

have 3 valance number

3 0
2 years ago
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Br2(l) and I-(aq) Use the reduction po
Vikki [24]

Answer:

1.58×10E18

Explanation:

Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.

E°cell= 1.07- 0.53= 0.54 V

E°cell= 0.0592/n logK

0.54 = 0.0592/2 logK

logK= 0.54/0.0296

logK= 18.2

K= Antilog (18.2)

K= 1.58×10^18

3 0
3 years ago
What observation would you look for to tell if a reaction was endothermic?
horsena [70]

Exothermic reactions release heat into their surroundings and endothermic reactions absorb heat and are cool. I believe the correct answer is D.

5 0
2 years ago
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