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2 years ago

Why does liquid move like it does when heated in the middle?

Chemistry

1 answer:

Lapatulllka [165]2 years ago

6 0

Answer:

heating makes the alcohol molecules of the liquid move faster.

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Identify the correct statements associated with titration.

Montano1993 [528]

Answer:

Explanation:

because titration by definition, is used to determine the concentration of one reactant in an aqueous solution, it requires a burette to accurately measure the volume of a solution and endpoints in acid-base titration can be determined by a colour indicator or a pH meter

3 0

3 years ago

Write the chemical equation for cellular respiration

BaLLatris [955]

Answer:

C 6 H 12 O 6 + 6O 2 → 6CO 2 + 6H 2 O + 38 ATP

Explanation:

7 0

2 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio

koban [17]

Answer:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

Explanation:

In this case, we can start with the formula of Platinum (II) Chloride. The cation is the atom at the left of the name (in this case $Pt^+^2$ ) and the anion is the atom at the right of the name (in this case $Cl^-$ ). With this in mind, the formula would be $PtCl_2$ .