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frosja888 [35]
3 years ago
11

Ethylene glycol (antifreeze) has a density of 1.11 g/cm3. What is the mass in grams of 385 mL of this liquid?

Chemistry
1 answer:
Makovka662 [10]3 years ago
6 0
Cm^3 = mL

1.11 g/cm^3 = 1.11 g/mL

Density (g/mL) multiplied by volume (mL) will give us the mass (g)

1.11 g/mL * 385 mL = 427.35 g

And since we have 3 significant figures, we have 427. g.
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3 years ago
What type of forest would you find south of spruce
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3 years ago
A 20 L cylinder of gas at 5.6 atm pressure at 35 degrees Celsius is heated to 95 degree Celsius and compared to 23 atm. What is
ella [17]

Answer:

5.8 L  

Step-by-step explanation:

This looks like a case where we can use the <em>Combined Gas Law</em> to calculate the temperature.

p₁V₁/T₁ = p₂V₂/T₂               Multiply both sides by T₂

p₁V₁T₂/T₁ = p₂V₂                Divide each side by V₂

V₂ = V₁ × p₁/p₂ × T₂/T₁  

=====

<em>Data</em>:

p₁ = 5.6 atm

V₁ = 20 L

T₁ = 35 °C = 308.15 K

p₂ = 23 atm

V₂ = ?

T₂ = 95 °C = 368.15 K

=====

<em>Calculation: </em>

V₂ = 20 × 5.6/23 × 368.15/308.15  

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6 0
3 years ago
Please help due in 5 min. I need this to be done or my grade will drop from a 96 to a 32.
ratelena [41]

Answer:

E = 3.035× 10-¹⁹J = 1.9eV

f = 4.58 × 10¹⁴Hz

Explanation:

wavelength = 6.55 × 10-⁷m

c = 3 × 10⁸m/s

f = ?

E = ?

a) f = c/wavelength

f = 3 × 10⁸/6.55 × 10-⁷

f = 4.58 × 10¹⁴Hz

b) E = hc/wavelength

E = 6.626×10-³⁴ × 3 × 10⁸/ 6.55 × 10-⁷

E = 3.035 × 10-¹⁹J

1ev = 1.6 × 10-¹⁹J

Therefore E = 3.035/1.6 = 1.9eV

3 0
3 years ago
Read 2 more answers
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