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lyudmila [28]
3 years ago
13

How many grams of AgNO3 are needed to prepare a 75.00 mL solution that is 6.5% (m/v)?

Chemistry
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

Volume of AgNO₃ = 4.9 ml (Approx)

Explanation:

Given:

Total solution = 75 ml

Volume of AgNO₃ = 6.5%

Find:

Volume of AgNO₃

Computation:

Volume of AgNO₃ = Total solution x Volume of AgNO₃

Volume of AgNO₃ = 75 x 6.5%

Volume of AgNO₃ = 4.875

Volume of AgNO₃ = 4.9 ml (Approx)

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35.0g of ice at 0 degrees

Explanation:

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How long does primary succession take? ty sm ily
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3 years ago
Calculate the boiling point of a 3.60 m aqueous sucrose solution. express the boiling point in degrees celsius to five significa
kvv77 [185]
We can use this equation for boiling point elevation:
ΔT(b) = i K(b) M
when Δ T(b) is the increase of boiling point of the solution.
and i is ( vant Hoff factor, the number of particles or ions per mole-clue.
and K(b) is boiling point increase constant for the solution ( and for water it is equal 0.52 C° Kg/mol)
We can assume i (vant Hoff factor ) = 1 as the sucrose is nonelectrolyte (not readily ionize).
So for water: Tb° = 100 c° and Kb = 0.52 c° Kg / mol
By substitute at:
ΔTb = i Kb M
∴  = 1 * 0.52 * 3.60 = 1.8432 C°
and when Tb = Tb° + ΔTb
∴  Tb = 100 + 1.8432 = 101.8432 C°



8 0
3 years ago
Read 2 more answers
A 12 gram piece of Cu at 475 oC is placed in contact with a 15 gram piece of Cr at 265 oC.
ale4655 [162]

Answer:

349.22°C

Explanation:

Let the final temperature of the two pieces of metal be x.

Now, the warmer metal which is C u reduces from 475°C to x. Thus Δt for C u is; Δt1 = 475 - x.

The cooler metal Cr increases in temperature from 265°C to x. Thus, it's change in temperature is Δt for Cr is; Δt2 = x - 265.

Now from conservation of energy, the amount of energy leaving the C u metal is equal to the amount of energy entering the Cr metal.

Thus;

q_lost = q_gain

Where;

q_lost = m1•c1•Δt1

q_gained = m2•c2•Δt2

Now, c1 & c2 are the specific heat capacity of C u and Cr respectively.

From online tables, c1 = 0.385 J/g°C and c2 = 0.46 J/g°C

We are given;

m1 = 12g and m2 = 15g

Thus;

12 × 0.385 × (475 - x) = 15 × 0.46 × (x - 265)

2194.5 - 4.62x = 6.9x - 1828.5

6.9x + 4.62x = 2194.5 + 1828.5

11.52x = 4023

x = 4023/11.52

x = 349.22°C

4 0
3 years ago
Which of the following is a good definition of matter?
yKpoI14uk [10]

Answer:

B

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I did the question before and got it right.

5 0
3 years ago
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