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Novosadov [1.4K]
3 years ago
14

Based on the activity series provided, which reactants will form products? F > Cl > Br > I CuI2 + Br2 Right arrow. Cl2

+ AlF3 Right arrow. Br2 + NaCl Right arrow. CuF2 + I2 Right arrow.
Chemistry
2 answers:
Sonja [21]3 years ago
8 0

Answer:

i believe the answer is a

Explanation:

agasfer [191]3 years ago
7 0

Answer: Cul2 + Br2 ->

Explanation:

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If one of the reactants in a reaction is Na20, what is known about the products?
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Answer:D

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Water contrasts on freezing ?
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No. When water first begins to cool down, it contracts. However, as it gets colder and eventually freezes, it begins to expand.

You can test this by freezing water in a water bottle: when you take it out of the freezer, the cap might have popped off or cracks may have formed in the sides of the bottle.

Answer: Water expands when frozen, not contracts.

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What are the blanks the missing blanks?
mixas84 [53]

Answer:

Cr: Mass- 51.9961 u

Fe: Moles- 55.845

Ti: Mass- 47.867 u

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3 years ago
3. Five grams of orange-drink mix are
navik [9.2K]

Answer:

I think it is C 105

Explanation:

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5 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
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