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2 years ago

How long does it take for a ketchup to dry on clothes

Chemistry

1 answer:

dexar [7]2 years ago

8 0

I'm going to say about 1-2 hours until it will completely dry. It depends on the material of the clothing.

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Calculate the mean rate of reaction when 23 cm<br> of gas<br> is produced in 15 seconds.

katrin2010 [14]

Answer:

1.53 cm/s

Explanation:

Rate = Length/time

= 23/15

= 1.53 cm/s

<u>BRAINLIEST?</u>

<u>PRETTY SURE IT IS.</u>

3 0

2 years ago

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

2 years ago

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

6 0

3 years ago

What has to be true of a substance for it to dissolve in water?

lord [1]

The polarity of a water molecule comes from the uneven distribution of electron density of hydrogen and oxygen atom. The oxygen in the water molecule is more electronegative than the hydrogen. Water has a partial positive charge near the hydrogen atom and a partial negative charge near the oxygen atom. The result of this electrostatic attraction results in the bond called hydrogen bond. Also, because of this bond, it has the ability to dissolve most of the solutes due to its polarity and bonding.

7 0

3 years ago

Which of the following intermolecular forces would affect boiling point the most?

Alekssandra [29.7K]

Hydrogen Bonding will effect the boiling point the most. Let's take an example Butane a four carbon unsaturated organic compound with molecular formula C₄H₁₀ and boiling point -1 °C.

H₃C-CH₂-CH₂-CH₃

Now, replace one hydrogen on terminal carbon with -OH group and convert it into Butanol.

H₃C-CH₂-CH₂-CH₂-OH

The Boiling point of Butanol is 117.7 °C. This increase in boiling point is due to formation of hydrogen bondings between the molecules of Butanol.

5 0

2 years ago