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jeka57 [31]
4 years ago
9

How many grams of ethylene glycol (C2H6O2) must be added to 1.15 kg of water to produce a solution that freezes at -4.46°C?

Chemistry
1 answer:
jok3333 [9.3K]4 years ago
7 0

Answer:

232.5 g C2H6O2

Explanation:

The equation you need to use here is ΔTf = i Kf m

Since pure water freezes at 0 C, your ΔTf is just 4.46 C

i = 1 (ethylene glycol is a weak electrolyte)

Kf = molal freezing constant, which for water is 1.86 C/m

m = molality = x mols C2H6O2 / 1.15 kg H2O (don't know the moles of ethylene glycol we're dissolving yet)

Than,

4.46 C = 1.86 C/m (x mol C2H6O2 / 1.15 kg H2O)

Solve for x, you should get x = 2.75 mol C2H6O2

3.75 mol C2H6O2 (62 g C2H6O2 / 1 mol C2H6O2) = 232.5 g C2H6O2

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Answer:

Above the Curie temperature, a magnet permanently loses all or some of its magnetism. External magnetic fields: Strong, opposing magnetic fields can cause the magnetic domains to lose their orientation and relax into a lower state of energy where they are not aligned.

Explanation:

4 0
3 years ago
I need help with number 5 please??
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8 0
3 years ago
An element and an atom are different but related because
White raven [17]

Answer:

c) An element is made up of all the same type of atom

Explanation:

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4 0
3 years ago
If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years
Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually, radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g

8 0
4 years ago
If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of
Ghella [55]

Answer:

The correct answer is - 0.570 grams

Explanation:

The balanced chemical reaction is given by

Cu(NO3)2(aq)     + 2NaOH(aq)    -------->    Cu(OH)2(s)      + 2NaNO3(aq)

        1.0 mole            2.0 mole                 1.0 mole          2.0 mole

number of mol of Cu(OH)2,

n = Molarity * Volume

= 35.0*0.167 = 5.845 millimoles

As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2

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= 97.5*5.845*10^-3

= 0.570 grams

Thus, the correct answer is - 0.570 grams

8 0
3 years ago
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