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3 years ago

Why are being flammable and combustible considered to be chemical properties and not physical properties?

Chemistry

1 answer:

g100num [7]3 years ago

6 0

Answer:

C. They change the substance into another substance.

Explanation:

Hope it helps!!

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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.25×10−4, what is the

Sergio [31]

Answer:

$1.2\times 10^{-2}$

Explanation:

The given reactions are:

PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq) $K_3 = 1.89\times 10^{-10}$

AgCl(aq)⇌Ag+(aq)+Cl−(aq) $K_4 = 1.25\times 10^{-4}$

Required reaction is:

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

$K_f = \frac{K_3}{K_4^2}\\=\frac{1.89\times 10^{-10}}{1.25\times 10^{-4}}^2\\=1.2\times 10^{-2}$

4 0

4 years ago

Molecule: Br2 and Br2. <br> Is it polar or nonpolar?

Ostrovityanka [42]

This combination in non polar.

7 0

3 years ago

Which catalyst is better for the decomposition of Hydrogen Peroxide(H₂O₂) , Potassium Iodide(KI) or Yeast? Explain why.

Yakvenalex [24]

Yeast, because it acts as a catalyst, or a helper, to remove the oxygen from the hydrogen peroxide

3 0

4 years ago

I NEED HELP PLEASE, THANKS!

Crazy boy [7]

Answer:

Here's what I get.

Explanation:

1. Brønsted-Lowry theory

An acid is a substance that can donate a proton to another substance.

A base is a substance that can accept a proton from another substance.

2. pH of ammonia

The chemical equation is

$\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \text{OH}$^{-}$$

For simplicity, let's re-write this as

$\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$$

(a) Set up an ICE table.

B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹: 0.335 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.335 + x x x

$\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.335}{1.8 \times 10^{-5}} = 28 000 > 400\\\\x \ll 0.335$

(b) Solve for [OH⁻]

$\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}$

(c) Calculate the pOH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.46 \times 10^{-3}) = 2.61$

(d) Calculate the pH

pH = 14.00 - pOH = 14.00 - 2.61 = 11.39

3 0

4 years ago

Use the Ref An aqueous solution of chromium(II) acetate has a concentration of 0.260 molal. The percent by mass of chromium(II)

poizon [28]

Answer:

The percent by mass of chromium(II) acetate in the solution is 4.42%.

Explanation:

Molality of the chromium(II) acetate solution = 0.260 m = 0.260 mol/kg

This means that in 1 kg of solution 0.260 moles of chromium(II) acetate are present.

1000 g = 1 kg

So, in 1000 grams of solution 0.260 moles of chromium(II) acetate are present.

Then in 100 grams of solution = $\frac{0.260 mol}{1000}\times 100=0.0260 mol$

Mass of 0.0260 moles chromium(II) acetate:

= 0.0260 mol × 170 g/mol = 4.42 g

$(w/w)\%=\frac{\text{mass of solute in 100 gram solution}}{100}\times 100$

$=\frac{4.42 g}{100 g}\times 100=4.42\%$

The percent by mass of chromium(II) acetate in the solution is 4.42%.

7 0

3 years ago