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nordsb [41]
3 years ago
6

Answer ian rlly kno ...........

Chemistry
2 answers:
stich3 [128]3 years ago
6 0
The slope would be -1/2
0-2
2+2 (cause one was negative add instead of subtract)
You would get -2/4 simply that to
-1/2
VMariaS [17]3 years ago
5 0
A ..............................
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BF3 acts as a Lewis acid when it accepts the lone pair of electrons that NH3 donates.
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Mashutka [201]

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7 0
3 years ago
Calculate the pressure (in kpa) of 1.5 mole of helium gas at 354 k when it occupies a volume of 16.5l.
3241004551 [841]

Answer:

267.57 kPa

Explanation:

Ideal gas law:

PV = n RT        R = 8.314462    L-kPa/K-mol

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2 years ago
The force of 20 N acts upon a 5kg block. What is the acceleration of the block?
charle [14.2K]

Answer:

<h2>The answer is 4 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

a is the acceleration

f is the force

m is the mass

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f = 20 N

m = 5 kg

We have

a =  \frac{20}{5}  \\

We have the final answer as

<h3>4 m/s²</h3>

Hope this helps you

8 0
3 years ago
Consider the reaction of CaCN2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3 . How m
adelina 88 [10]

Answer:

56 g. Option 3.

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The reaction is: CaCN₂ + 3H₂O → CaCO₃ + 2 NH₃

1 mol of calcium cianide reacts with 3 moles of water in order to produce 1 mol of calcium carbonate and 2 moles of ammonia

We have the mass of each reactant, so let's convert the mass to moles:

45 g. 1mol / 80.08 g = 0.562 moles of cianide

45 g. 1mol / 18 g = 2.5 moles of water

The cianide is the limiting reactant:

3 moles of water need 1 mol of cianide to react

Then, 2.5 moles of water will need (2.5 . 1)/ 3 = 0.833 moles

As we have 0.562 moles of CN⁻ we don't have enough

We can work now, on the reaction:

Ratio is 1:1. Therefore 0.562 moles of cianide will produce 0.562 moles of carbonate

Let's convert the mass to moles to find the answer:

0.562 mol . 100.08 g / 1 mol = 56.2 g

8 0
2 years ago
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