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3 years ago

Patrick and SpongeBob love to blow bubbles! Patrick found some Super Bubble Soap at Sail-Mart. The ads

Chemistry

1 answer:

EleoNora [17]3 years ago

4 0

Answer:

BUBBLE SUPER BUBBLE

Explanation:

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Which of these is a pure substance

Galina-37 [17]

Both A and B are the answer.

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3 years ago

Binomial nomenclature example and sentence using that word<br><br> please help im going to cry

Alexandra [31]

What’s wrong? ...........:.........

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3 years ago

The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co

garri49 [273]

Answer:

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater ( $m_{sw}$ ), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

$m_{sw} = \rho_{sw}\cdot V_{sw}$

Where:

$\rho_{sw}$ - Density of seawater, measured in grams per liters.

$V_{sw}$ - Volume of seawater, measured in liters.

If $V_{sw} = 1.5\times 10^{21}\,L$ and $\rho_{sw} = 1030\,\frac{g}{L}$ , then:

$m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)$

$m_{sw} = 1.545\times 10^{24}\,g$

The total mass of sodium chloride is determined by the following ratio:

$r = \frac{m_{NaCl}}{m_{sw}}$

$m_{NaCl} = r\cdot m_{sw}$

Given that $m_{sw} = 1.545\times 10^{24}\,g$ and $r = 0.035$ , the total mass of sodium chloride in all the seawater in the world is:

$m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)$

$m_{NaCl} = 5.408\times 10^{22}\,g$

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

8 0

3 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

<u>Where:</u>

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

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3 years ago

do ionic compounds or covalent compounds generally have weaker forces of attraction between the ions or molecules making up the

Crank

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2 years ago