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Debora [2.8K]
3 years ago
5

Why is water so susceptible to pollution?(1 point)

Chemistry
1 answer:
seropon [69]3 years ago
8 0
A. It’s known as the “universal solvent”
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Number of
Karolina [17]
Mass plus atomic and number

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3 years ago
What is the mass of 3.62X10^24 molecules of methanol
Bad White [126]
N(CH₃OH)=3,62·10²⁴/6·10²³ 1/mol = 6,033 mol
m(CH₃OH) = 6,033 mol · 32 g/mol (molar mass) = 193,06 g.
7 0
3 years ago
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The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions
mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0
3 years ago
What mass of water could be warmed from 21.4 degrees celsius to 43.4 degrees celsius by the pellet dropped inside it? Heat capac
Artist 52 [7]

42.34 g of water could be warmed from 21.4°C to 43.4°C  by the pellet dropped inside it

Heat loss by the pellet is equal to the Heat gained by the water.

q_{w} = -q_{p} ….(1)

where, q_{w} is the heat gained by water

q_{p} is the heat loss by pellet

q_{w} = mCΔT

where m = mass of water

C = specific heat capacity of water = 4.184 J/g-°C

ΔT = Increase in temperature

ΔT for water = 43.4 - 21.4 = 22°C

q_{w} = m × 4.184 × 22 …. (2)

Now

q_{p} = H_{c} ×ΔT

where H_{c} = Heat capacity of pellet = 56J/°C

Δ T for pellet = 43.4 - 113 =- 69.6°C

q_{p} = 56 × -69.6 = -3897.6 J

From equation (1) and (2)

-m× 4.184 × 22 =-3897.6

m= 42.34 g

Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.

Learn more about specific heat here brainly.com/question/16559442

#SPJ1

6 0
2 years ago
When you knock on a door, what causes you to hear a sound?
I am Lyosha [343]

Answer: B.

Explanation:

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