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Nadusha1986 [10]
2 years ago
14

What to do when bored and have no electronics?

Chemistry
2 answers:
eduard2 years ago
4 0
Workout play basketball play cards
Solnce55 [7]2 years ago
3 0

Answer:

Explanation:

1. Get a pillow

2. Find a bed or a couch

3. Get a comfy blanket

4. SLEEP

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Which statement BEST describes what happens during solvation?
Ymorist [56]

Answer:

A

Explanation:

3 0
2 years ago
Help please please help please
Anni [7]

Answer:

I don't fully understand what this is about...

Explanation:

sorry :(

5 0
3 years ago
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To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber
Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}

7 0
3 years ago
If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio
bulgar [2K]

Answer:

m_{Cd(OH)_2}=36.6 gCd(OH)_2

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

4 0
2 years ago
Which statement is correct? A. pKa is not an indicator of acid strength. B. An acid with a small Ka is stronger than an acid wit
Svetradugi [14.3K]

Answer:

D

Explanation:

Since [pKa = - log Ka]....hence..,the larger the Ka value,the stronger the acid is..so this means that the pKa is vice versa

Saying that the smaller the pKa value..the stronger the acid is.

3 0
3 years ago
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