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3 years ago

Use your understanding of the ideal gas law to

2 answers:

8 0

Answer:pressure is inversely related to volume. Temperature is inversely related to moles. Volume is inversely related to pressure. Moles are inversely related to temperature.

PolarNik [594]3 years ago

4 0

Answer:The ideal gas law is represented mathematically as: PV=nRT. P- pressure, V- volume, n-number of moles of gas, R- ideal gas constant, T- temperature.

Explanation:The ideal gas law is used as a prediction of the behavior of many gases, when subjected to different conditions.

he ideal gas law has so many limitations.

An increase in the pressure or volume, decreases the number of moles and temperature of the gas.

Empirical laws that led to generation of the ideal gas laws, considered two variables and keeping the others constant. This empirical laws include, Boyle's law, Charles's law, Gay Lusaac's law and Avogadro's law.

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09. Where does Mitosis take place? I

mash [69]

Answer:

In plants mitosis takes place throughout life in growing regions called the meristems. Replacements as cells wear out. The cells of the skin and bone marrow are sites of active mitosis replacing skin cells and red blood cells that only have a limited life. Repair.

Explanation:

5 0

3 years ago

a normal hemoglobin concentration in the blood is 15g/100ml of blood how many kilogram of hemoglobin are in a person who has 5.5

jok3333 [9.3K]

Using stoichiometry:

5.5 L of blood x (1000 mL/1L) x (15 g/100 mL) x (1 kg/1000 g) = 0.825 kg

4 0

3 years ago

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A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r

Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

5 0

2 years ago

Why is water in the gas phase when it comes out of a volcano?

umka2103 [35]

Due to lava which heats water where water vapour is comes

4 0

3 years ago

Read 2 more answers

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

Read 2 more answers