In a lab experiment, john uses a mesh to separate soil particles from water. Which technique of separation is he using

In a chemical reaction, 36 grams of hydrochloric acid reacts with 40 grams of sodium hydroxide to produce sodium chloride and wa

Elena L [17]

The answer is <span>The mass of sodium chloride formed is less than 76 grams.
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Remember Mass is never gained or loss in any chemical reactions

So,

hydrochloric acid + sodium hydroxide => <span>sodium chloride + water
Total 76 gram Total will also 76 gram

In this case, the mass of sodium chloride will be less than 76 gram because we need to take out the mass of the water.

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3 0

4 years ago

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A topographic map would be most useful for which activity?

Ainat [17]

the answer is D. because a topographic map measures the lowest and highest points in a certain place

Hope this helps

5 0

3 years ago

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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:

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

3 0

3 years ago

The equation for the combustion of CH4 (the main component of natural gas) is

Lera25 [3.4K]

Heat produced = -13588.956 kJ

<h3>Further explanation</h3>

Given

The reaction of combustion of Methane

CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔH∘rxn=−802.3kJ

271 g of CH4

Required

Heat produced

Solution

mol of 271 g CH₄ (MW=16 g/mol0

mol = mass : MW

mol = 271 : 16

mol = 16.9375

So Heat produced :

= mol x ΔH°rxn

= 16.9375 mol x −802.3kJ/mol = -13588.956 kJ

6 0

3 years ago

How does the law of conservation of mass apply to this reaction C2H4+4O2-->4H2O+2CO2 .Only the oxygen needs to be balanced B.

r-ruslan [8.4K]

C2H4+4O2-->4H2O+2CO2

the answer is C

6 0

3 years ago

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