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3 years ago

Please help. Last grade of this class

Chemistry

1 answer:

Marat540 [252]3 years ago

5 0

Answer: Question 3

Group 14, Period 3 is a metalloid

Explanation: Question 3

The element in group 14, period 3 on the periodic table is Si. Metalloids are elements which share both non-metallic and metallic characteristics.

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If you are given an ideal gas with pressure (p)259,392.00 pa and temperature (T)=200°c of 1 mole Argon gas in a volume 8.8dm3,ca

GuDViN [60]

Answer: $R=4.82436 \frac{Pa. m^{3}}{mol. K}$

Explanation:

The Ideal Gas equation is:

$P.V=n.R.T$ (1)

Where:

$P$ is the pressure of the gas

$n$ the number of moles of gas

$R=8.3144598 \frac{Pa. m^{3}}{mol. K}$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin.

$V$ is the volume

It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that an ideal gas is a single hypothetical gas. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be $8.3144598 \frac{Pa. m^{3}}{mol. K}$ .

However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.

Having this clarified, let's isolate $R$ from (1):

$R=\frac{PV}{nT}$ (2)

Where:

$P=259392 Pa$

$n=1 mole$

$T=200\°C=473.15 K$ is the absolute temperature of the gas in Kelvin.

$V=8.8 dm^{3}=0.0088 m^{3}$

$R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}$ (3)

Finally:

$R=4.82436 \frac{Pa. m^{3}}{mol. K}$

4 0

3 years ago

Help please thank you!!!

omeli [17]

Answer:

I think its J

Explanation:

sorry if im wrong

7 0

2 years ago

Read 2 more answers

Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO2H) and 0.230 mol of sodium formate (NaCO2H)

Veronika [31]

The question is incomplete, here is the complete question:

Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴

a) 2.099

b) 10.463

c) 3.546

d) 2.307

e) 3.952

Answer: The pH of the solution is 3.546

Explanation:

We are given:

Moles of formic acid = 0.370 moles

Moles of sodium formate = 0.230 moles

Volume of solution = 1 L

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})$

$pK_a$ = negative logarithm of acid dissociation constant of formic acid = 3.75

$[HCOONa]=\frac{0.230}{1}$

$[HCOOH]=\frac{0.370}{1}$

pH = ?

Putting values in above equation, we get:

$pH=3.75+\log(\frac{0.23/1}{0.37/1})\\\\pH=3.54$

Hence, the pH of the solution is 3.546

5 0

3 years ago

Things that are too small (or too large) to see can be studied with models

MakcuM [25]

Answer:

Things that are too small (or too large) to see can be studied with models

True

8 0

2 years ago

Read 2 more answers

Answer A B C and D please for my chemistry hw

r-ruslan [8.4K]

Answer: The equations are provided below.

Explanation:

Skeleton equations are defined as the equations which simply indicate the molecules that are involved in a chemical reaction. These equations are unbalanced equations.

Balanced equations are defined as the chemical equation in which number of individual atoms on the reactant side must be equal to the number of individual atoms on the product side.