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myrzilka [38]
2 years ago
13

What two major Southeastern U.S. cities will be underwater if sea level rose 6 feet?

Chemistry
1 answer:
miss Akunina [59]2 years ago
6 0

Answer:

Rising sea levels would be devastating in North Somerset, with large parts of Clevedon, Weston-super-Mare and Burnham all being underwater by 2050 according to Climate Central

Explanation:

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Cloud condensation nuclei or CCNs are small particles typically 0.2 µm, or 1/100 the size of a cloud droplet on which water vapor condenses. Water requires a non-gaseous surface to make the transition from a vapour to a liquid; this process is called condensation.

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2 years ago
PLEASE I NEED HELP ASAP
Murljashka [212]

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Because the density of water is one

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3 years ago
Fe + H2SO4 = Fe2(SO4)3 + H2
a_sh-v [17]
C (667) that’s the answer boiii
5 0
2 years ago
please help me with Chem I ONLY HAVE 5 MINUTES if methane gas (CH4) flows at a rate of 0.25L/s, how many grams of methane gas wi
Nookie1986 [14]

Answer:

643g of methane will there be in the room

Explanation:

To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:

<em>Volume Methane:</em>

3600s * (0.25L / s) = 900L Methane

<em>Moles methane:</em>

PV = nRT; PV / RT = n

<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>

Replacing:

PV / RT = n

1atm*900L / 0.082atmL/molK*273.15 = n

n = 40.18mol methane

<em>Mass methane:</em>

40.18 moles * (16g/mol) =

<h3>643g of methane will there be in the room</h3>
5 0
3 years ago
In a neutralization reaction, 24.6 mL of 0.300 M H2SO4(aq) reacts completely with 20.0 mL of NaOH(aq). The products are Na2SO4(a
stiv31 [10]

Answer : The concentration of the NaOH solution is, 0.738 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL

Putting values in above equation, we get:

2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL

M_2=0.738M

Thus, the concentration of the NaOH solution is, 0.738 M

7 0
3 years ago
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