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3 years ago

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What do you know about a substance it you know its temperature? A.How heavy the substance is B.How large or small the particles

of the substance are C.How many particles are in the substance D.How hot or cold the substance is

2 answers:

Softa [21]3 years ago

4 0

Answer:

C. how many particles are in the substance

Hitman42 [59]3 years ago

3 0

Answer:

it is how hot or cold so bassically c in my test

Explanation:

You might be interested in

Calculate the molar mass of RbOH

Ghella [55]

Hey there!

RbOH

Rb: 1 x 85.468 = 85.468

O: 1 x 16 = 16

H: 1 x 1.008 = 1.008

------------------------------------

102.476

The molar mass of RbOH is 102.476 g/mol.

Hope this helps!

5 0

3 years ago

What best describes the goals of scientific investigation and technological development?

bekas [8.4K]

Hi! I think the answer is option B. I hope
this helps, Goodluck :)

5 0

3 years ago

Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a

fiasKO [112]

Answer:

SEPARATION SCHEME FOR CATIONS

GIVEN CATIONS : $Ag^{+} \ , Fe^{3+} , Cu^{2+}, Ni^{2+}$

Step 1: Add $6mol/dm^3$ of $HCl$ to the mixture solution

Result : This would cause a precipitate of $AgCl$ to be formed

Reaction : $Ag^{+} _{(aq)} + Cl^{-} _{(aq)} ---------> AgCl(ppt)$

Step 2 : Next is to remove the precipitate and add $H_2S$ to the remaining

solution in the presence of $0.2 \ mol/dm^3$ of HCl

Result : This would cause a precipitate of $CuS$ to be formed

Reaction : $Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)$

Step 3: Next remove the precipitate then add $6 \ mol/dm^3$ of aqueous

$NH_3 (NH_3 \cdot H_2 O)$ , process the solution in a centrifuge,when the

process is done then sort out the precipitate from the solution

Now this precipitate is $Fe(OH)_3$ and the remaining solution

contains $(Ni (NH_3)_6)$

Next take out the precipitate to a different beaker and add HCl

to it this will dissolve it, then add a drop of $NH_4SCN$ this will

form a precipitate $Fe(SCN)_{6}^{3-}$ which will have the color of

blood indicating the presence of $Fe^{3+}$

Reaction : $F^{3+}_{(aq)} + 30H^-_{(aq)} --------->Fe(OH)_3_{(aq)}$

$Fe (OH)_{(s)} _3 + 3H^{+}_{aq} -------> Fe^{3+}_{aq} + 3H_2O_{(l)}$

$Fe^{3+} + 6SCN^{-} -----> Fe(SCN)_6 ^{3-}$

Now the remaining mixture contains $Ni^{2+}$

Explanation:

6 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

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3 years ago

During which process of the water cycle do water molecules absorb a great amount of solar energy? A. evaporation B. condensation

jolli1 [7]

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3 years ago