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sergij07 [2.7K]
3 years ago
15

How much water needs to be added

Chemistry
1 answer:
Olegator [25]3 years ago
8 0

Answer:

165 ml

Explanation:

We are given;

Initial volume; V_a = 55 ml

Initial molarity; M_a = 3 M

Molarity of desired solution; M_b = 0.75 M

Volume of desired solution; V_b = (55 + x) ml

Where x is the volume of water to be added.

To solve for V_b, we will use the equation ;

M_a•V_a = M_b•V_b

V_b = (M_a•V_a)/M_b

V_b = (3 × 55)/0.75

V_b = 220 mL

Thus;

(55 + x) = 220

x = 220 - 55

x = 165 mL

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<h3>Name of the elements</h3>

The names of the elements that make up the compound will be determined from the atomic mass of the elements.

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Thus, when sulfur reacts with zinc metal is forms a compound known as zinc sulfide.

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1 year ago
How much heat is required to raise the temperature of 81.0 g of water from its melting point to its boiling point?
Dovator [93]

Answer:

Specific heat of water = 33.89 KJ

Explanation:

Given:

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Q = Mass x Specific heat of water x (Final temperature - Initial temperature)

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6 0
3 years ago
Does the identity of gas matter when predicting its behavior why or why not?
just olya [345]

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3 years ago
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer
lidiya [134]

The question is incomplete, the complete question is

Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V

Answer:

only IV and V

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If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.

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7 0
3 years ago
I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:
dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

2.  Use the equation P₁V₁ = P₂V₂.  Solve for V₂.  

(3.05 L)(870 kPa) = (969 kPa)(V₂)  

V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

(3.32 L)/(360 K) = (4.50 L)/(T₂)  

T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

V₂ = 7.47 L

5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)  

V₂ = 38.6 L

6.  Do the same as above, but for n₂.

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n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

6 0
3 years ago
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