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Anna11 [10]
3 years ago
13

Pleas help fill in blank

Chemistry
1 answer:
Nana76 [90]3 years ago
3 0
First on is transport
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Why boron trifluoride is nonpolar and sulphur dioxide is polar
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it small because that bond is not very polar notice that all of those arrows they cancel. because they point in opposite directions. that's why BH 3 is nonpolar if we were to draw the Lewis structure of bf3. ... so even though the boron fluorine bond is polar the molecule as a whole is nonpolar.

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Is Bromothymol Blue acidic or basic?
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What two things must all matter have?
OleMash [197]

1.) Mass

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3 years ago
Read 2 more answers
Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to
hram777 [196]

Answer:

0.1056 mole

Explanation:

As Sally knows that the charge on the metal ion is n = +2

$MCl_n=MCl_2$

In that compartment $[M^{n+}]=[m^{2+}]=8.279 \ M$

The volume of the $MCl_n$ taken in that compartment = 6.380 mL

So, the number of moles of $M^{2+} = 8.279 \times 6.380$

                                                      = 52.82 m mol

                                                      = 0.05280 mol

$MCl_n \rightarrow M^{n+}+nCl^-$

But n = 2

Therefore, moles of $Cl^-$ = 2 x moles of $M^{n+}$

                                       = 2 x 0.05282

                                       = 0.1056 mole

3 0
3 years ago
5. How much heat (in calories) is absorbed by a reaction when
Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

  • Mass of water (m): 300 g
  • Initial temperature: 80 °C
  • Final temperature: 40°C

We can calculate the heat released by the water (Q_w) when it cools using the following expression.

Q_w = c \times m \times (T_f - T_i)

where

c is the specific heat capacity of water (1 cal/g.°C)

Q_w = \frac{1cal}{g.\°C}  \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal

According to the law of conservation of energy, the sum of the heat released by the water (Q_w) and the heat absorbed by the reaction (Q_r) is zero.

Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal

7 0
2 years ago
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