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laila [671]
4 years ago
15

A significant limitation for early scientists studying the elements was

Chemistry
1 answer:
liq [111]4 years ago
7 0
<span>a. Lack of precise measuring instruments.</span>
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How many kilometers can the car travel on the amount of gasoline that would fit in a soda can? The volume of a soda can is 355 m
g100num [7]
I'll assume that the car is a Honda Insight with <span>EPA gas mileage rating of 57mi/gal in the city.

First, we will convert all units into Km and mL:
1 mile = </span><span>1.609344 Km
</span><span>1gal = 3785.411 mL
</span>
Then, we will calculate the distance the car can move by multiplying the mileage rating and the the volume available as follows (note that I will be converting units in the same step):
distance = <span>(57mi/gal) x (1.609344 km/1mi) x (1gal/3785.411mL) x 355mL 
               = 8.6027 Km</span>
4 0
4 years ago
Read 2 more answers
Starting from position 3,5 go west 7 , then south 7, then east 4 what’s your ending point?
Vsevolod [243]

Answer:

-1, -2

Explanation:

West is left south is down east is right

5 0
3 years ago
Please delete this question. Solved!
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6 0
3 years ago
Some properties, like boiling point elevation and freezing point depression, change in solutions due to the presence and number
Reil [10]

D. quantitative properties

Boiling point and freezing point depression are both values that can be represented quantitatively (in number form).

5 0
3 years ago
Read 2 more answers
At 66.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so
andreev551 [17]

Answer:

\boxed{5.5 \times 10^{-28}}

Explanation:

We must use the Nernst equation

E = E^{\circ} - \frac{RT}{zF}\ln Q

Step 1. Calculate E°

SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ ⇌ SO₂(g) + 2H₂O(ℓ)

<u>2Br⁻⇌ Br₂(aq) + 2e⁻                                                                </u>

SO₄²⁻(aq) + 4H⁺(aq) + 2Br⁻(aq) ⇌ Br₂(aq) + SO₂(g) + 2H₂O(ℓ)

E° = 0.17 - 1.0873 = -0.92 V

Step 2. Calculate Q

E   =  0      V

E° = -0.92 V

R = 8.314 J·K⁻¹mol⁻¹

T = 66 °C

n = 2

F = 96 485 C/mol

Calculations:  

T = 66.0 + 273.15 = 339.15 K

0 = -0.92 - \dfrac{8.314 \times 339.15}{2 \times 96 485}\ln Q\\\\0.92 = -0.0146\times \ln Q \\\\\ln Q = -\dfrac{0.92}{0.0146} = -62.8\\\\Q = e^{-62.8} =\boxed{5.5 \times 10^{-28}}\\

6 0
4 years ago
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