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3 years ago

Label the reactants, products, and intermediates.

Chemistry

1 answer:

Annette [7]3 years ago

5 0

Of what?i cant see any pic here?

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PLEASE PLEASE HELP ME!!!A 450 ml gas sample has a pressure of 1.25 atm at 65 °C. What is the temperature, in °C, at which the ga

Arlecino [84]

Answer:

89°C

Explanation:

Combined Gas Law (P₁V₁)/T₁ = (P₂V₂)/T₂

(1.25 atm)(450 mL)/(65°C) = (0.89 atm)(865 mL)/T₂

8.653846154 = 769.85/T₂

T₂ = 769.85/8.653846154

T₂ = 88.96044444 = 89°C

6 0

2 years ago

A gas occupies a volume of 2.45 L at a pressure of 1.03 atm. What volume will the

Gemiola [76]

Answer:

2.58 L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

5 0

3 years ago

Write the difference between the reaction of Calcium and Aluminium with water.

guapka [62]

Answer:

Both gives hydroxides, but

Calcium reacts with water to give calcium hydroxide Ca(OH)2. this product is slightly soluble in water.

Aluminium is less reactive than Ca, so it reacts with only steam, to give aluminium hydroxide Al(OH)3. This product is insoluble in water.

5 0

2 years ago

How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

$pH+pOH=14\\pOH=14-8.94=5.06$

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

Now we have to calculate the moles of $OH^-$ .

Formula used : $Moles=Concentration\times Volume$

$\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles$

For neutralization, equal number of moles of $H^+$ ions will neutralize same number of $OH^-$ ions.

$\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles$

As, $H_2SO_4\rightarrow 2H^++SO^{2-}_4$

From this reaction, we conclude that

2 moles of $H^+$ ion is given by the 1 mole of $H_2SO_4$

$3309.8\times 10^{-9}$ moles of $H^+$ ion is given by $\frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9}$ moles of $H_2SO_4$

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of $H_2SO_4$ × Molar mass of sulfuric acid

Mass of sulfuric acid = $(1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g$

Therefore, the mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

3 0

2 years ago

In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o

Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

$V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1} )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w$