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fgiga [73]
3 years ago
13

Smog is a type of air pollution. It is a mixture of fog and smoke or other Airborne pollutants such as exhaust fumes. What is a

result of small that affect the quality of life for organisms?
Chemistry
1 answer:
Tanya [424]3 years ago
8 0
The answer is acid rain 

for study island your welcome :)
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16.
mixas84 [53]

Answer:

Explanation:

When going from top to bottom in any group of the periodic table, the atomic radius always tends to increase

8 0
2 years ago
Between which two elements is the difference in metallic character the greatest?
xz_007 [3.2K]

Rb and O is your answer

6 0
3 years ago
Should we only write condensed formula in chemical equations (for lessons like carbon and compounds). What did your teachers say
Firlakuza [10]

Answer:

Here's what I get.

Explanation:

  • If your teachers don't ask for a specific type of formula, a condensed structural formula should be OK.
  • If they ask specifically for a structural formula or a bond-line formula, that is what you must give.

Bottom line: ask your teachers in advance what they expect.

7 0
3 years ago
Calculate the pH of a solution that is 0.210 M in nitrous acid (HNO2) and 0.290 M in potassium nitrite (KNO2). The acid dissocia
jeka94

Answer:

pH = 3.49

Explanation:

We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH  of a buffer ssytem using the Henderson-Hasselbach equation.

pH = pKa + log [base] / [acid]

pH = -log Ka + log [NO₂⁻] / [HNO₂]

pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M

pH = 3.49

6 0
3 years ago
An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the
sattari [20]

Answer:

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

P_i = 100 N·m

T_i = 327 K

P_f = 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln(P_f/P_i) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈<em> -0.52571</em> kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln(P_f/P_i) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = <em>-7.5 kJ</em>

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = <em>-5.00574 kJ</em>

6 0
2 years ago
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