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2 years ago

The melting point of oxygen is -218°C and its boiling point is -183°C. What is the state of oxygen at -200°C?Immersive Reader

Chemistry

1 answer:

Anna007 [38]2 years ago

3 0

it is in a liquid state

Explanation:

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J5 protons and 74 electrons

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Answer:

ins the same

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3 years ago

Need to complete the chart

GalinKa [24]

Answer:

Row 1

$[H^+]=1.8\times 10^{-6}M$

$pH=-\log[H^+]=-\log[1.8\times 10^{-6}]=5.7$

pOh=14-pH=14-5.7=8.3

$pOH=-\log[OH^-]$

$[OH^-]=0.5\times 10^{-8}M$

Hence, acidic

Row 2

$[OH^-]=3.6\times 10^{-10}M$

$pOH=-\log[OH^-]=-\log[3.6\times 10^{-10}]=9.4$

pH=14-pOH=14 - 9.4 = 4.6

$pH=-\log[H^+]$

$[H^+]=2.6\times 10^{-5}M$

Hence, acidic

Row 3

pH = 8.15

$[H^+]=0.7\times 10^{-8}M$

pOH=14-pH=14 - 8.15 = 5.8

$pOH=-\log[OH^-]$

$[OH^-]=1.5\times 10^{-6}M$

Hence, basic

Row 4

pOH = 5.70

$[OH^-]=1.8\times 10^{-6}M$

pH=14-pOH=14 - 5.70 = 8.3

$pH=-\log[H^+]$

$[H^+]=0.5\times 10^{-8}M$

Hence, basic

3 0

3 years ago

A student was given a stock solution with a concentration of 3.61 g/mL and asked to perform

vovangra [49]

Answer:

The concentration of the dilute sample will be 0.361 g/ml

Explanation:

If a solution is diluted into 1:10 ratio then the amount of solute of that solution will be decreased by 10 times.

The initial concentration of the stock solution was 3.61g/ml but when the solution is diluted in 1:10 ratio the solute concentration is also decreased by 10 times.SO at present the solute concentration becomes 3.61/10=0.361 g/ml.

8 0

3 years ago

A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay fo

Ksenya-84 [330]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$80 = x*20$

$80 = 20\:x$

$20\:x = 80$

$x = \dfrac{80}{20}$

$\boxed{x = 4}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{43}{2^{4}}$

$m = \dfrac{43}{16}$

$\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

8 0

3 years ago