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3 years ago

Need help fast!!!!!!!!!!!

Chemistry

1 answer:

icang [17]3 years ago

3 0

I think it’s the second one!! Hope this helps! BRANLIEST plzzzz

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If 26.2 grams of a pure compound contain 8.77 × 1022 molecules, what is the molecular weight of this compound? Answer in units o

Nataly [62]

Answer:

Mw = 179.845 g/mol

Explanation:

Mw [=] g/mol

∴ w = 26.2 g

∴ 1 mol = 6.02 E23 molecules.......Avogadro's number

⇒N° moles = 8.77 E22 molecules * ( mol / 6.02 E23 molecules ) = 0.146 mol

⇒ Mw = 26.2 g / 0.146 mol = 179.845 g/mol

3 0

3 years ago

lead can react with oxygen gas. If lead (IV) oxide is the product of the reaction, how would the reaction be classified

slamgirl [31]

Answer:

this reaction is an oxidation reaction

7 0

3 years ago

5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?

Aleksandr [31]

Answer:

$V_2=9.20L$

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

$\frac{V_2}{n_2}= \frac{V_1}{n_1}$

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

$V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L$

Best regards!

3 0

3 years ago

How many molecules of SrCrO4 are in a sample of SrCrO4 0.556 moles?

Amiraneli [1.4K]

Answer:

$3.35*10^{23}\ SrCrO_4\ molecules$

Explanation:

$We\ are\ given\ that,\\No.\ of\ moles\ of\ SrCrO_4=0.556\\Hence,\\As\ we\ know\ that,\\No.\ of\ particles=Avagadro's\ Constant*No.\ of\ moles\\We\ already\ know\ that\ Avagadro's\ Constant=6.022*10^{23}\\Here,\\No.\ of\ SrCrO_4\ molecules= 6.022*10^{23}*0.556\\Hence,\\No.\ of\ SrCrO_4\ molecules=3.348*10^{23} molecules\ \approx 3.35*10^{23}\ SrCrO_4\ molecules$

7 0

3 years ago

A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is:

Luden [163]

Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have

$64 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 100 X

B) total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $500 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $20 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 178.88 X

8 0

3 years ago