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3 years ago

A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?

1 answer:

3 0

STP means standard temperature and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the numer of moles and can be represented as mass of the gas, m, divided by the molar mass, c. so we have,

PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P

substitute the values into the equation,
c = [(0.63g/L)(0.08206 L-atm/mol-K)(273K)]/(1atm)
c = 14.11 g/mol

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What is the specific heat of a substance if 1560 cal are required to raise the temperature of a 312-g sample by 15°C?

Oksanka [162]

The specific heat capacity the substance is calculated using the below formula

Q(heat) = Mc delta T
Q =1560 cal
m(mass) 312 g
delta T (change in temperature ) = 15 c
C= specific heat capacity=?

by making c the subject of the formula
c=Q/m delta T
= 1560 cal/ 312g x 15 c = 0.33 cal/g/c (answer B)

8 0

3 years ago

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0

OleMash [197]

Answer: The mass of original oxalic acid sample is 6.75 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams

7 0

3 years ago

5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0

3 years ago

For the quantum number lower case L = 0, how many possible values are there for the quantum number m subscript l?

liubo4ka [24]

Only 0 is possible for ml

6 0

4 years ago

A sample of gas occupies 0.04 L at 150 K. What volume does that sample occupy at 300 K?

Natalka [10]

C.
hope this helps :)

8 0

3 years ago