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3 years ago

10 points. Please help.

Chemistry

2 answers:

lana66690 [7]3 years ago

8 0

Answer:

The temperature is 81.3 K.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

In this case:

P= 2.5 atm
V= 8 L
n=3 moles
R= 0.082 $\frac{atm*L}{mol*K}$
T= ?

Replacing:

2.5 atm* 8 L= 3 moles* 0.082 $\frac{atm*L}{mol*K}$ *T

Solving:

$T=\frac{2.5 atm* 8 L}{3 moles* 0.082\frac{atm*L}{mol*K}}$

T= 81.3 K

The temperature is 81.3 K.

ratelena [41]3 years ago

6 0

Answer:

-191.7°C

Explanation:

P . V = n . R . T

That's the Ideal Gases Law. It can be useful to solve the question.

We replace data:

2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°

(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°

T° = 81.3 K

We convert T° from K to C°

81.3K - 273 = -191.7°C

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Convert 947ml to Liters

vitfil [10]

Answer:

.947

Explanation

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3 years ago

Can some body please help me with this Stoichiometry stuff

andriy [413]

Answer:

See explanations

Explanation:

Stoichiometry is very easy to master if you understand the ‘mole concept’ and how it is used to define and describe chemical process mathematically. A ‘mole’ – in chemistry – is the mass of substance containing one Avogadro’s Number of particles. That is, N₀ = 6.023 x 10²³ particles / mole. When working with chemical reactions and equations data should be first converted to moles using the following conversations:

1 mole = 1 formula weight = 6.023 x 10²³ particles = 22.4 liters at STP(0⁰, 1atm).

In this problem you are given the equation Na + H₂O => NaOH + H₂. ‘Reading the equation’ there is 1 mole of Na, 1 mole of water, 1 mole of NaOH and 1 mole of H₂. In another example 3H₂ + N₂ => 2NH₃ there are 3 moles of H₂, 1 mole of N₂ and 2 moles of NH₃. The mole values can be multiples or fractions but if one mole value increases all the remaining mole values increase or decrease proportionally. For example:

Using the equation Na + H₂O => NaOH + H₂, one could apply a 2 before the Na but all the following formulas would need be increased by a factor of 2. If one applies ½ to the Na then all the following formulas would need be cut in half also and the reaction stoichiometry would still be valid. The fact that the equation is written with coefficients of 1 is that it is in the smallest whole number ratio of coefficients. This then implies the reaction formula is in ‘standard form’. This also implies the equation conditions are at 0⁰C & 1atm pressure and 1 mole of any gas phase substance occupies 22.4 Liters volume. Such is the significance of converting given data to moles as all other substance mass (in moles) are proportional.

For your 1st problem, 1.76 x 10²⁴ formula units of Na will react with water (usually read as an excess) to produce (?) grams of H₂.

1st write the equation followed by listing the givens below the respective formulas… That is…

Na + H₂O => NaOH + H₂,

Given: 1.76 x 10²⁴ atoms excess --------- ? grams

Convert atoms Na to moles = 1.76 x 10²⁴atoms/6.023 x 10²³atoms/mole

=2.922moles Na produces=>2.922moles H₂(because moles Na=moles H₂).

Convert moles to grams =>2.922moles H₂ x 2.000 grams H₂/mole H₂

=5.8443 grams H₂

2nd problem, 3.5 moles Na will react with H₂O (in excess) to produce (?) moles of NaOH.

Again write equation and assign values to each formula unit in the equation.

Na + H₂O => NaOH + H₂,

Given: 3.5moles excess ? grams ----

Since coefficients of balanced std equation are equal then moles Na equals moles of NaOH, that is, 3.5 moles Na produces => 3.5 moles NaOH

Convert moles NaOH to grams => 3.5 moles NaOH x 40 g NaOH/mole NaOH = 140 grams NaOH

3rd problem, 2.75 x 10²⁵ molecules H₂O will react with (?) atoms of Na.

Same procedure, convert to moles, solve problem by ratios then convert to needed dimension at end of problem.

Na + H₂O => NaOH + H₂

Given: ? atoms 2.75 x 10²⁵ molecules H₂O => NaOH + H₂

Convert to moles => 2.75 x 10²⁵ molecules H₂O / 6.023 x 10²³ molecules H₂O/mole H₂O = 45.658 moles H₂O => 45.658 moles Na (equal coefficients)

Convert moles Na to atoms Na => 45.658 moles Na x 6.023 x 10²³atoms Na/mole Na = 2.75 x 10²⁵ atoms Na.

Note => Problem 3 could have been solved by inspection b/c coefficients are equal, however, always go through a process that you can justify and defend even if it does take longer. Never assume anything. Depend on what you know, not what you 'think' you know.

Master the mole concept and you master a lot of chemistry! Good luck.

5 0

3 years ago

How many grams of H3PO4 are in 300 mL of a .50 M solution of H3PO4

Liono4ka [1.6K]

The first thing you need to do is convert mL into L
so (175 mL)(.001L/1mL)=.175L then you multiply by the Molarity of H3PO4 which is 3.5mol/L so (.175L)(3.5mol/L)=.6125 mol H3PO4, and since it wants the answer in grams you then multiply (.6125molH3PO4) by the molar mass of H3P04 which is about 97.99g and your answer is 60.02g which is about 60 grams of H3PO4. Hope this helped

3 0

4 years ago

Using a spectrophotometer, and a cuvette with a path length of 1 cm you measure the absorbance (A275) of Guanosine to be 0.70. C

Anton [14]

Answer : The concentration of guanosine in your sample is, $8.33\times 10^{-5}M$

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution = 0.70

C = concentration of solution = ?

l = path length = 1.00 cm

$\epsilon$ = molar absorptivity coefficient guanosine = $8400M^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$0.70=8400M^{-1}cm^{-1}\times C\times 1.00cm$

$C=8.33\times 10^{-5}M$

Thus, the concentration of guanosine in your sample is, $8.33\times 10^{-5}M$

5 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

sergeinik [125]

The question is incomplete, here is the complete question:

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Answer: The percent yield of water in the reaction is 46.85 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaOH:

Given mass of NaOH = 77.0 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{77.0g}{40g/mol}=1.925mol$

For sulfuric acid:

Given mass of sulfuric acid = 72.6 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

So, 0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid produces 2 moles of water

So, 0.741 moles of sulfuric acid will produce = $\frac{2}{1}\times 0.741=1.482moles$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

$1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.68g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 12.5 g

Theoretical yield of water = 26.68 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{12.5g}{26.68g}\times 100\\\\\% \text{yield of water}=46.85\%$

Hence, the percent yield of water in the reaction is 46.85 %.

8 0

3 years ago