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love history [14]
2 years ago
9

Identify the Brønsted–Lowry acid and the Brønsted–Lowry

Chemistry
1 answer:
Margaret [11]2 years ago
4 0

The following applies to the above equation:

  • Brønsted–Lowry acid - HCOOH
  • Brønsted–Lowry base - PO43-
  • Conjugate acid - HPO42-
  • Conjugate base - HCOO-

<h3>What is Bronsted-Lowry acid and base?</h3>

Brønsted-Lowry acid is the any chemical species that acts as a donor of protons while Brønsted-Lowry base is any chemical species that acts as a proton acceptor.

In the following equation: HCOOH(aq) + PO43-(aq) = HCOO-(aq) + HPO42-(aq)

  1. HCOOH is the proton (H+) donor, hence, is the Brønsted-Lowry acid
  2. PO43- is the donor acceptor, hence, is the Brønsted-Lowry base
  3. HCOO- is the conjugate base that forms from the acid
  4. HPO42- is the conjugate acid that forms from the base

Learn more about Brønsted-Lowry acid and base at: brainly.com/question/15885173

#SPJ1

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A particle of mass 2.0 kg moves under the influence of the force F(x)=(-5x^2+7x)~\text{N}F(x)=(−5x ​2 ​​ +7x) N. If its speed at
Brums [2.3K]

Answer:

The speed of the particle at x = 4.0 m is 13.66 m/s

Explanation:

The work done by this force between the two points above is given by

W = ∫ F dx

W = ∫⁴₋₄ (-5x² + 7x) dx

W = [(-5x³/3) + (7x²/2)]⁴₋₄

W = [(-5(4³)/3) + (7(4²)/2)] - [(-5(-4)³)/3) + (7((-4)²)/2)] = (-50.6667) - (162.6667) = (- 213.33 J)

Kinetic energy at -4.0 m

At this point, v = 20 m/s

K.E = mv²/2 = 2 × 20²/2 = 400 J

To obtain the kinetic energy at 4 m,

We apply the work-energy theorem which mathematically translates to

The work done in moving a particle from one point to another = Change in kinetic energy of the particle between those two points

W = ΔK.E

Work done between x = - 4m and x = 4 m is - 213.33 J

Hence, ΔK.E = -213.33 J

Change in kinetic energy of the particle between x = - 4m and x = 4m is ΔK.E

ΔK.E = (Kinetic energy at x = 4m) - (kinetic energy at x = - 4m)

- 213.33 J = (mv²/2) - 400

mv²/2 = -213.33 + 400 = 186.67 J

2v² = 2 × 186.67

v² = 186.67

v = 13.66 m/s.

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a 360mg sample of sugar (molar mass of 180 g/mol) is dissolved in enough water to produce 200mL of solution. after the sugar is
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