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2 years ago

1. The definition of the relative atomic mass for an element, X, is:

Chemistry

1 answer:

sweet-ann [11.9K]2 years ago

7 0

Nolur acil lütfen yalvarırım sana da

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Does a substance that heats up quickly have a high or a low specific heat capacity

Oliga [24]

A low specific heat capacity

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3 years ago

Physical Science pre-Test 2020-2021

poizon [28]

Answer:

exothermic I think

Explanation:

IM TWKIGN THE TEST

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2 years ago

You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the

prisoha [69]

Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

<em>Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.</em>

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

That means the moles of NaOH added = Moles of sodium citrate produced

And the resulitng moles of HX = Initial moles - Moles NaOH added

<em>Moles HX and NaX:</em>

Moles NaOH = 0.100L * (0.65mol / L) = 0.065 moles NaOH = Moles NaX

Moles HX = 0.300L * (0.45mol / L) = 0.135 moles HX - 0.065 moles NaOH = 0.070 moles HX

Replacing in H-H equation:

pH = 3.14 + log [0.065mol] / [0.070mol]

pH = 3.11 is the pH of the buffer

8 0

2 years ago

If 125 gram of 75% pure Caco3 is treated with 100 gram of Hcl to produce Cacl2 H20 and Co2 Find which one is limiting reactant

damaskus [11]

Answer:

Explanation:

7 0

2 years ago

For the reaction IO3–(aq) + 5 I–(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) the rate of disappearance of I–(aq) at a particular time a

scoray [572]

Answer: The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$IO_3^-(aq)+5I^-(aq)+6H^+(aq)\rightarrow 3I_2(aq)+3H_2O(l)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}$

Given: $-\frac{d[I^-]}{dt}]$ = $2.4\times 10^{-3}mol/Ls$

$+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls$

The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

6 0

3 years ago