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3 years ago

What will be the observations for the reaction: CuSO4(aq) + NaOH(aq)? Its a two mark question so two observations at least.

1 answer:

6 0

The chemical reaction would be:

<span>CuSO4(aq) + NaOH(aq) = Cu(OH)2(s) + Na2SO4(aq)

One observation would be the formation of a precipitate since one of the products is not readily soluble to aqueous solution. A formation of a blue precipitate will be observed.</span>

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Please refer to image please

GarryVolchara [31]

Answer:

76.9L

Explanation:

Based on the graph, whenever the temperature increases by 100K, the volume increases by 10L, so do 769/10

7 0

2 years ago

a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is

fiasKO [112]

Answer: The empirical formula is $CH_2$ .

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.142}{0.142}=1$

For H = $\frac{0.287}{0.142}=2$

The ratio of C: H = 1: 2

Hence the empirical formula is $CH_2$ .

4 0

3 years ago

Which have more molecules ? 20 g of ammonia or 30 g of oxygen gas?

hodyreva [135]

I think it's oxygen gas

5 0

2 years ago

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

8 0

3 years ago

Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation o

Korolek [52]

The rate of disappearance of chlorine gas : 0.2 mol/dm³

<h3>Further explanation</h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

$\tt aA+bB\rightarrow cC+dD$

The rate reaction :

$\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}$

Reaction for formation CCl₄ :

<em />

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of CCl₄ = reaction rate x coefficient of CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of Cl₂ = reaction rate x coefficient of Cl₂

Rate of disappearance of Cl₂ = 0.05 x 4 = 0.2 mol/dm³

4 0

3 years ago