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pshichka [43]
3 years ago
11

Hi I like a guy (He knows I like him too cause my friend told him) but me and this guy are really good friends how do I deal wit

h this we both get all awkward when around each other now and he can barely get a sentence out now and I turn as red as a tomatoe.....
(I is a girl too)
What do I do please help. I would also like a guys perspective on this pleeeeeeeeeeaaaaaase!!!!!!!!!!!!
Chemistry
1 answer:
kirza4 [7]3 years ago
8 0

Answer:

Explanation:

When you start to feel you are desiring more than just a casual friendship with your best friend and you’re not sure what to do next, let me offer you some advice that could strengthen both your relationship and your love for each other.

First, don’t rush into a romantic relationship with your best friend…many times people confuse love with that other kind of caring love you feel for all of your other friends.

Second, don’t spill your guts right away. You might feel like you have to share all your thoughts and feelings with the other person as soon as you start to feel something. That’s usually a mistake.

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Determine how many liters of hydrogen adjusted to STP there are in a 50.0 liter steel cylinder if the pressure inside is 100.0 a
Dvinal [7]

Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 100.0 atm

P_2 = final pressure of gas at STP = 1 atm

V_1 = initial volume of gas = 50.0 L

V_2 = final volume of gas at STP = ?

T_1 = initial temperature of gas = 27.0^oC=273+27.0=300K

T_2 = final temperature of gas at STP = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}

V_2=4550L

Therefore, the volume of hydrogen gas at STP is 4550 L.

3 0
3 years ago
"A crosslinked copolymer consists of 57 wt% ethylene (C2H4) repeat units and 43 wt% propylene (C3H6) repeat units. Determine the
ch4aika [34]

Explanation:

Percentage ethylene by weight = 57%

Percentage propylene by weight = 43%

Suppose in 100 grams of polymer:

Mass of ethylene = 57 g

Mass of propylene = 43 g

Moles of ethylene = \frac{57 g}{28 g/mol}=2.036 mol

Moles of propylene = \frac{43g}{42g/mol}=1.024 mol

1 mole = N_A =6.022\times 10^{23} molecules/ atoms

Units of ethylene = 2.036 mol\times N_A

Units of propylene = 1.024 mol\times N_A

a) Fraction of ethylene units:

=\frac{2.036 mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{509}{765}

b ) Fraction of propylene units:

=\frac{1.024mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{256}{765}

3 0
3 years ago
Assuming equal concentrations, rank these solutions by ph.? assuming equal concentrations, rank these solutions by ph. koh (aq h
polet [3.4K]
I remember coming across this question and the options were:
KOH, HCN, NH₃, HI, Sr(OH)₂

Now, a substance with a low pH is one that dissociates completely in water to release hydrogen ions, while basic substances dissociate completely to release hydroxide ions. Therefore, in the order of increasing pH:
HI, HCN, NH₃, Sr(OH)₂, KOH
7 0
3 years ago
I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti
Tanya [424]
<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
4 0
3 years ago
A buffer consists of 0.45 M CH3COOH (acetic acid) and 0.35 M CH3COONa. The Ka of acetic acid is 1.8 x 10-5 . a) Calculate the pH
Zigmanuir [339]

Answer:

A) pH of Buffer solution = 4.59

B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original    buffer solution = 4.65

Explanation:

This  is the Henderson-Hasselbalch Equation:

 pH = pKa + log\frac{[conjugate base]}{[acid]}

to calculate the pH of the following Buffer solutions.

8 0
4 years ago
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