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3 years ago

5.80 gg of NaCl in 350. mL of a NaClsolution

Chemistry

1 answer:

Delicious77 [7]3 years ago

6 0

Answer:

Solution is 0.28 M

You can also say, [NaCl] = 0.28 mol/L

Explanation:

As you have a solute mass and the solution's volume, you may find the molarity concentration of solution.

Molarity specifies the moles of solute in 1 L of solution

We convert the volume of solution to L → 350 mL . 1L / 1000 mL = 0.350L

We convert the mass of solute to moles → 5.80 g . 1mol / 58.45 g = 0.0992 moles

Molarity (mol/L) = 0.0992 mol /0.350L = 0.28M

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How many moles of gas sample are 5.0 L container at 373K and 203kPa

Rom4ik [11]

For the purpose we will here use the ideal gas law:

p×V=n×R×T

V= 5.0 L

T= 373K

p= 203kPa

R is universal gas constant, and its value is 8.314 J/mol×K

Now when we have all necessary date we can calculate the number of moles:

n=p×V/R×T

n= 203 x 5 / 8.314 x 373 = 0.33 mole

6 0

3 years ago

Helppppppppppppppppppppp

alukav5142 [94]

decameters - meters: multiply by 10

meters to meters: multiply by 1

centimeters to meters: divide by 100

millimeters to meters: divide by 1000

For the rows at the bottom:

hectometer row: 100, multiply by 100, 4500

decameter row: 10, multiply by 10, 450

meter row: 1, multiply by 1, 45

decimeter row: 0.1, divide by 10, 4.5

centimeter row: 0.01, divide by 100, 0.45

im guessing theres a millimeter row at the bottom:

millimeter row: 0.001, divide by 1000, 0.045

hope this helps!

5 0

3 years ago

Balance the equation αC6H14(g)+βO2(g)→γCO2(g)+δH2O(g)

fenix001 [56]

Answer: The balanced reaction is:
2 C6H14(g)+ 19 O2(g) → 12 CO2(g)+ 14 H2O(g)

Note: While balancing the chemical reaction, care must be taken that total number atoms (of each type) on both reactant and product side must be same. In present case, there are 12 'C' atoms, 28 'H' atoms and 38 'O' atoms on both reactant and product side. Hence, the reaction is balanced.

7 0

3 years ago

Read 2 more answers

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

Baking soda and antacid tablets are examples of two common__________

attashe74 [19]

Bases

A base is a substance that dissociates into more hydroxide ions (-OH-) when dissolved in water. Bases are also good proton acceptors. Bases, therefore, reduce the number of H+ and increase OH- hence raising the pH of the solution.

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

Explanation:

Other properties of bases is that they are bitter to the taste and they feel slippery when touched. Strong bases are nonthlese very corrosive like acids. Bases turn red litmus paper blue. Most alkali hydroxides such as NaOH are bases.

Learn More:

For more on bases check out;

brainly.com/question/12574229

brainly.com/question/2015251

#LearnWithBrainly

7 0

3 years ago