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QveST [7]
2 years ago
10

HCl reacts with Barium Hydroxide to produce barium chloride and water. how many ml of a 3.00 M hydrochloric acid solution would

be required to react with 25.5 ml of a 4.65 M barium hydroxide solution?
Chemistry
1 answer:
ollegr [7]2 years ago
3 0
THINK OF GUMMY BEARS IN A TEST TUBE AND IN THAT TEST TUBE THEIRS A SUBSTANCE ACID IT EX PLODS LOOK IT UP ON YOU TUBE
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Sodium (chemical symbol Na) is atomic number 11 and is in the first column of the periodic table so it has one valence electron.
lisabon 2012 [21]
Na is cation so it loses electron to be positive and become stable losing one valence shells one electron so it's oxidation number is +1 ie A is correct
7 0
2 years ago
Read 2 more answers
Why would it be better to be an r-selected species if the water resources in an area were to become more limited over a short pe
Gnoma [55]
An r-selected species reproduces much faster than K-selected species.

r-selected species focuses on maturing and reproducing quickly. r-selected species will probably reproduce when the water supply is there for the short period of time; thus, increasing the chance of the r-selected species of surviving.

K-selected species, on the other hand, focus on raising their young and reproduce later. Since the K-selected species take long to mature before reproducing, water supply may run out before they have a chance of fully maturing; thus, K-selected species have a lower chance of survival.

Hope this helps.
If you need anything more, feel free to comment! Have an awesome day! :)

~Collinjun0827, Junior Moderator
6 0
1 year ago
Read 2 more answers
Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole
icang [17]

Answer:

Explanation has been given below.

Explanation:

  • Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
  • Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
  • First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
  • Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
  • Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
  • Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
3 0
2 years ago
Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d
PilotLPTM [1.2K]

Answer:

k = 23045 N/m

Explanation:

To find the spring constant, you take into account the maximum elastic potential energy that the spring can support. The kinetic energy of the car must be, at least, equal to elastic potential energy of the spring when it is compressed to its limit. Then, you have:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the car = 1050 kg

k: spring constant = ?

v: velocity of the car = 8 km/h

x: maximum compression of the spring = 1.5 cm = 0.015m

You solve the equation (1) for k. But first you convert the velocity v to m/s:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant is 23045 N/m

3 0
1 year ago
The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is __________.
crimeas [40]
PH is calculated using <span>Handerson- Hasselbalch equation,

                                      pH  =  pKa  +  log [conjugate base] / [acid]

Conjugate Base  =  Acetate (CH</span>₃COO⁻)
Acid                    =  Acetic acid (CH₃COOH)
So,
                                      pH  =  pKa  +  log [acetate] / [acetic acid]

We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

                                     pKa  =  -log Ka
Putting value of Ka,
                                     pKa  =  -log 1.76 × 10⁻⁵

                                     pKa  =  4.75
Now,
Putting all values in eq. 1,
                     
                                     pH  =  4.75 + log [0.172] / [0.818]

                                     pH  =  4.072
4 0
2 years ago
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