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valentina_108 [34]
3 years ago
15

How many grams are there in 457.25 pounds?

Chemistry
2 answers:
Usimov [2.4K]3 years ago
8 0

there are 2.2 lbs /kg. take 457.25/2.2=207.8409091
there are 1000g/kg so take your answer times 1000= 207840.9091g

Ne4ueva [31]3 years ago
7 0
There are 207405.111 grams in that many pounds.
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Dmitri Mendeleev is the answer

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Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s
slava [35]

Answer:

Potassium carbonate (K₂CO₃)

Explanation:

The compounds dissociate into ions in water, as follows:

K₂CO₃ → 2 K⁺ + CO₃⁻    ⇒ 3 dissolved particles per mole

NaI → Na⁺ + I⁻    ⇒ 2 dissolved particles per mole

KBr → K⁺ + Br⁻   ⇒ 2 dissolved particles per mole

CH₃OH → CH₃O⁻ + H⁺  ⇒ 2 dissolved particles per mole

NH₄Cl → NH₄⁺ + Cl⁻   ⇒ 2 dissolved particles per mole

Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).

3 0
3 years ago
50 examples word equation with balanced chemical equarion​
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5 0
1 year ago
Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80
Damm [24]

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³      (1l= 1000 cm³)

                            = 1010 g

% (by mass) = 48.00 g/ 1010 g  x 100 = 4.8 %

6 0
3 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0
3 years ago
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