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3 years ago

How does the smell of food being cooked spread so fast ???

Chemistry

2 answers:

natta225 [31]3 years ago

6 0

Answer:

smell of cooked food spread fast bc when something is heated up, the kinetic energy of the molecules/particles increases and some of them even mix with the air and moves around. the first answer is also correct (i just explained it more detaily)

so, this is the reason :)

mark as brainliest dagini ;)

sladkih [1.3K]3 years ago

5 0

Answer:

When we cook food in the kitchen, that's the region of higher concentration of the smell. By diffusion, the smell spreads to the whole room and thereby whole house, so anyone standing at a distance, can smell it.

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Which of these substances has the highest pOH? 0.10 M HCl, pH = 1 0.001 M HNO3, pH = 3 0.01 M NaOH, pH = 12 The answer is 0.10 M

Naya [18.7K]

Answer:On these combined scales of pH and pH it can be shown that because for water when pH = pH = 7 that pH + pH = 14. This relationship is useful in the inter conversion of values. For example, the pH at a 0.01 M solution of sodium hydroxide is 2, the pH of the same solution must be 14-2 = 12.

Explanation:

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3 years ago

Daltons idea that atoms cannot be divided into smaller parts was disproved by the discovery of the.

tensa zangetsu [6.8K]

It was disproved by the discovery of the electrons

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3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

What is state symbols

Ganezh [65]

Answer:

Explanation:

State symbols are used in chemical equations to delineate the state of matter in which the reaction is taking place.

They give a good perspective of the state of the reactants and products obtainable.

There are basically four states of matter in every chemical reaction:

Solids are symbolized by small letter (s)
Liquids are represented by (l)
Gases are shown by (g)
Aqueous solutions having water as the medium by (aq)

These symbols appear as subscript in front of the chemical species.

5 0

3 years ago

If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?

padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, $t_{(1/2)}$ = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

$N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }$

For two half lives, t = 2 × $t_{(1/2)}$

∴ t = 2 × 15 = 30

$\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2$

$\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75$

∴ The amount left, N(t) = 0.75 grams

4 0

3 years ago