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3 years ago

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Who got insta gramm because I need help with my science assignment I’ll pay u 30$ cashapp or PayPal if it’s right I need a good

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1 answer:

g100num [7]3 years ago

3 0

Okay yeah sure you don’t need to pay me tho

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When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef

kumpel [21]

Answer:

<em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

<em></em>

Explanation:

<em>mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.</em>

mass = 3.51 g

<em>lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec</em>

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

<em>From graham's equation of effusion</em>,

$\frac{Rc}{Rp}$ = $\sqrt{\frac{Mp\\}{Mc} }$

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

$\frac{3.51}{1.98}$ = $\sqrt{\frac{Mp\\}{44.01} }$

1.77 = $\frac{\sqrt{Mp} }{6.64}$

11.75 = $\sqrt{Mp}$

Mp = $11.75^{2}$

Mp = <em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

5 0

3 years ago

For the reaction2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g)ΔH° is -125 kJ/mol and ΔS° is +253 J/K ∙ mol. This reaction is _

ivanzaharov [21]

Answer:

C. spontaneous at all temperatures

Explanation:

The spontaneity of reaction is determined by the sign of the gibbs free energy.

A negative sign denotes that the reaction is spontaneous, positive sign means the reaction is not spontaneous.

From the question;

ΔS° = +253 J/K

ΔH° = -125 kJ/mol

ΔG = ΔH° - TΔS°

From the data given, the condition in which we can obtain a negative value of G, is at any value of T.

For any value of T, G would always be a negative value.

This means the correct option is option C.

3 0

3 years ago

Which member of each pair is more soluble in water? Why?<br> (a) CH₃CH₂OCH₂CH₃(l) or CH₃CH₂OCH₂(g)

Anna [14]

CH₃CH₂OCH₂ is more soluble in water because it has shorter hydrocarbon chain.

<h3>What is hydrocarbon?</h3>

Hydrocarbon is defined as the compound which contain hydrocarbon and carbon atoms.

The carbon atom attached to each other to form framework and hydrogen atom attach to them in different ways to give different configuration. One of the most popular hydrocarbon compound is diamond.

<h3>Solubility of hydrocarbon in water</h3>

Hydrocarbon is non polar compound whereas water is polar compound. So, hydrocarbon is in soluble in water. But as they have weak intermolecular interactions known as London dispersion forces i.e. Instantaneous dipole-induced dipole interactions.

make them less soluble in water.

Greater the hydrocarbon chain lesser will be the solubility of ketone in water. On the other hand, lesser the hydrocarbon chain greater will be the solubility of ketone in water.

Thus, we concluded that the CH₃CH₂OCH₂ is more soluble in water because it has shorter hydrocarbon chain.

learn more about hydrocarbon:

brainly.com/question/16020705

#SPJ4

7 0

2 years ago

If I add 2.75 g of CaO to 155 mL of distilled water, what will be the concentration of the solution be?

IRINA_888 [86]

M = m/L
moles of CaO = 2.75g / molar mass
2.75g/ 56.08g = 0.049moles CaO
155mL / 1000 = 0.155 L
M= 0.049moles/0.155L
M=0.316

7 0

2 years ago

Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho

AleksandrR [38]

The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
$(1/2) ^{n} =x$
$t_{1/2} = \frac{t}{n}$
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span> $t_{1/2}$ - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span> $(1/2) ^{n} =x$
</span> $(0.5) ^{n} =0.125$
$n*log(0.5)=log(0.125)$
$n= \frac{log(0.5)}{log(0.125)}$
$n=3$

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span> $t_{1/2} = 12.4$
</span><span> $t_{1/2} *n = t$
</span> $t= 12.4*3$

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>

7 0

4 years ago

Read 2 more answers