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2 years ago

Convert 150 K to degrees F

Chemistry

1 answer:

Ne4ueva [31]2 years ago

6 0

Answer:

150K is 253.67°F

Explanation:

There is a simple formula to convert from Kelvin to Celsuis;

C° = K - 273.15

Then you would convert Celsuis into Fahrenheit using the formula:

$F = \frac{9}{5}C + 32\\$

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The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.

Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions ${\text{IO}_{3}}^{-}$ .

How many moles of iodine $\text{I}_2$ will be produced?

${\text{IO}_{3}}^{-}$ converts to $\text{I}_2$ in the first reaction. The coefficient in front of $\text{I}_2$ is three times the coefficient in front of ${\text{IO}_{3}}^{-}$ . In other words, each mole of ${\text{IO}_{3}}^{-}$ will produce three moles of $\text{I}_2$ . 0.0020 moles of ${\text{IO}_{3}}^{-}$ will convert to 0.0060 moles of $\text{I}_2$ .

How many moles of thiosulfate ions ${\text{S}_2\text{O}_3}^{2-}$ are required?

$\text{I}_2$ reacts with ${\text{S}_2\text{O}_3}^{2-}$ in the second reaction. The coefficient in front of $\text{I}_2$ is twice the coefficient in front of ${\text{S}_2\text{O}_3}^{2-}$ . How many moles of ${\text{S}_2\text{O}_3}^{2-}$ does each mole of $\text{I}_2$ consume? Two. 0.0060 moles of $\text{I}_2$ will be produced. As a result, $2 \times 0.0060 = 0.0120$ moles of ${\text{S}_2\text{O}_3}^{2-}$ will be needed.

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3 years ago

How could you determine if a tissue you are looking at could only be from the heart?

borishaifa [10]

You can determine it by paying attention to the unique characteristics that could only be found at heart's tissue, such as :

- looks striated or stripped
- The bundles are breached like tree but connected at both ends

hope this helps

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3 years ago

Why do scientists use models to study atoms?

umka2103 [35]

Because atoms are the small and can't be seen by your eyes. It is so small that scientists need to use a model. A model help scientists study things. So, scientists needs to study atoms using models.

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3 years ago

how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C

andrezito [222]

Answer:

333.7g of antifreeze

Explanation:

Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:

ΔT = Kf × m × i

Where:

ΔT is change in temperature (0°C - -20°C = 20°C)

Kf is freezing point depression constant (1.86°C / m)

m is molality of solution (moles solute / 0.5 kg solvent -500g water-)

i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)

Replacing:

20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1

5.376 = moles solute

As molar mass of ethylene glycol is 62.07g/mol:

5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.

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3 years ago

How are half life and radioactive decay related

hoa [83]

Answer : Half life and radioactive decay are inversely proportional to each other.

Explanation :

The mathematic relationship between the half-life and radioactive decay :

$N=N_oe^{-\lambda t}$ ................(1)

where,

N = number of radioactive atoms at time, t

$N_o$ = number of radioactive atoms at the beginning when time is zero

e = Euler's constant = 2.17828

t = time

$\lambda$ = decay rate

when $t=t_{1/2}$ then the number of radioactive decay become half of the initial decay atom i.e $N=\frac{N_o}{2}$ .

Now substituting these conditions in above equation (1), we get

$\frac{N_o}{2}=N_oe^{-\lambda t_{1/2}}$

By rearranging the terms, we get

$\frac{1}{2}=e^{-\lambda t_{1/2}}$

Now taking natural log on both side,

$ln(\frac{1}{2})=-\lambda \times t_{1/2}$

By rearranging the terms, we get

$t_{1/2}=\frac{0.693}{\lambda}$

This is the relationship between the half-life and radioactive decay.

Hence, from this we conclude that the Half life and radioactive decay are inversely proportional to each other. That means faster the decay, shorter the half-life.

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3 years ago