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lina2011 [118]
3 years ago
11

What do you think is the effect of acid rain on rusting ?

Chemistry
1 answer:
stich3 [128]3 years ago
8 0
As acid rain is a mixture of pollutants and gasses. As rusting is a chemical reaction between oxygen and other gasses in the air alongside a material and a liquid. I would say they are both similar and that acid rain would speed the process of rusting/aid the process.
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Has anyone ever done this atom escape room thing
SVEN [57.7K]

Answer:

1.  negative

2.  positive

3.  neutral

Explanation:

Ok so it looks like they are asking for the charge (positive, negative, or neutral) of each thing

So for 1, it would be negative, because it's pointing to an electron.  Electrons always have a negative charge.

So for 2, it would be positive, because it's pointing to a proton.  Protons always have a positive charge

So for 3, it would be neutral, because it's pointing to a neutron.  Neutrons always have a neutral charge.

4 0
3 years ago
Is Columns of elements in the periodic table are called periods correct in the underlined sentence
SSSSS [86.1K]
I don't know sorry I really need point
6 0
3 years ago
Which of the following breaks and forms bonds?
svp [43]
Your answer would be "Chemical Change"
7 0
3 years ago
NEED ANSWER FAST!
lorasvet [3.4K]

Answer:

B

Explanation:

Molarity = 0.010M

Volume = 2.5L

Applying mole-concept,

0.010mole = 1L

X mole = 2.5L

X = (0.010 × 2.5) / 1

X = 0.025moles

0.025moles is present in 2.5L of NaOH solution.

Molar mass of NaOH = (23 + 16 + 1) = 40g/mol

Number of moles = mass / molar mass

Mass = number of moles × molar mass

Mass = 0.025 × 40

Mass = 1g

1g is present in 2.5L of NaOH solution

6 0
3 years ago
A 17.4 L sample of oxygen gas (O2) was collected at a temperature of 23.0°C and a pressure of 2.18 atmospheres. What volume woul
timofeeve [1]

The volume of the gas at STP = 35.01 L

<h3>Further explanation</h3>

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

V=17.4 L

T = 23 + 273 = 296 K

P = 2.18 atm

\tt mol=n=\dfrac{PV}{RT}\\\\n=\dfrac{2.18\times 17.4}{0.082\times 296}\\\\n=1.563

The volume of the gas occupy at STP :

\tt 1.563\times 22.4=35.01~L

6 0
3 years ago
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