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fiasKO [112]
3 years ago
11

Which of the following is a vapor? steamhidrógen oxygen helium​

Chemistry
1 answer:
Scrat [10]3 years ago
7 0
Helium bc when u vape it goes into it lungs but also bringing helium at the same time.
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1. Which of the following qualifies as a Bronsted-Lowry base but not an Arrhenius base?
Luden [163]

Arrhenius Theory: according to Arrhenius, acid is one that can donate proton in an aqueous solution, while base is one that can donate hydroxide ion in an aqueous solution.

Bronsted-Lowry Theory: according to Bronsted Lowry, acid is one that can donate protons while base is one that can accept a proton.

1. In first, only C. NH3 can't give hydroxide ion, but can accept a proton so it is a Bronsted-Lowry Base but not an Arrhenius base.

2.In second, as the definition suggested, bronsted base is one that can accept protons and acid is one that can loose protons. so answer is D. Acids lose H+ and bases gain H+.


5 0
3 years ago
Read 2 more answers
List the number of
rjkz [21]

Answer:

beryllium: 2 valence electrons

nitrogen: 5  valence electrons

oxygen: 6 valence electrons

fluorine: 7  valence electrons

magnesium: 2+ valence electrons

phosphorus: 5 valence electrons

sulfur: 6 valence electrons

chlorine: 7  valence electrons

Explanation:

6 0
3 years ago
You need a 70% alcohol solution. On hand, you have a 135 mL of a 20% alcohol mixture. You also have 85% alcohol mixture. How muc
frutty [35]

To obtain the desired solution:

450 mL of 85% alcohol solution is needed to obtain the desired solution.  

Calculation:

Let x be the amount of the 85% alcohol required

The volume of the resulting 70% alcohol solution will then be = x + 135 ml

135 mL of the 20% alcohol solution contains the amount of "pure" alcohol is =  0.20×135 mL.

The 85% alcohol solution contains x mL of "pure" alcohol = 0.85× x mL.

The total amount of the "pure" alcohol is the sum

=  0.20×135 + 0.85× x mL.

It should be equal to the amount of the "pure" alcohol in the mixture, which is = 0.70× (x+135) ml.

So, your "pure alcohol" equation is,  

=  0.85× x + 0.20×135 = 0.70× (x+135)

Simplify and solve it for x:

0.85x + 0.20×135 = 0.70x + 0.70×135,

0.85x - 0.70x = 0.70×135 - 0.20×135,

0.15x = 67.5

x = 67.5/0.15

= 450mL.

Learn more about the alcohol solution here,

brainly.com/question/12246176

#SPJ4

 

7 0
2 years ago
A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill
tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

N_{t} = N_{0}e^{-\lambda t}

Where:

λ: is the decay constant = ln(2)/t_{1/2}

t_{1/2}: is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg

Now, at 7:00 pm we have:

t = 7:00 pm - (5:00 pm + 90 min) = 30 min

N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg    

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0
3 years ago
How would you prepare 175 mL of a 0.350kmol/m^3 calcium nitrate solution
SVEN [57.7K]
0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol

mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol

1 mol --------- 164g
0,06125 ---- X
X = 10,045g

To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
3 0
3 years ago
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