All categories

2 years ago

how many grams of sodium hydroxide are required to dissolve in 232 g of water to make a 2.88 m solution

Chemistry

1 answer:

Gemiola [76]2 years ago

7 0

Grams of Sodium hydroxide : 26.72 g

<h3>Further explanation</h3>

Molality shows the number of moles dissolved in every 1000 grams of solvent.

m = n. (1000 / p)

m = Molality

n = Number of moles of solute

p = Solvent mass (gram)

Can be written :

$\tt m=\dfrac{mol~solute}{kg~solvent}$

mol solute(NaOH)

m=2.88

kg solvent(water)=232 g=0.232 kg

$\tt mol~solute=m\times kg~solvent\\\\mol~solute=2.88\times 0.232=0.668$

mass of NaOH

$\tt mass=mol\times MW\\\\mass=0.668\times 40~g/mol=26.72~g$

You might be interested in

For science i had to create a super hero from an element. i got beryllium.

kvasek [131]

Hello :)

You can spell it multiple ways depending on what you like best :)
Here are a few:

Resykae
Reisekae
Resiykay

Hope this helps!
btw, nice name choice :))

5 0

2 years ago

If two reactions sum to an overall reaction, and the equilibrium constants for the two reactions are K1K1 and K2K2, what is the

inessss [21]

Answer: The equilibrium constant for the overall reaction is $K_1\times K_2$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) $A\rightarrow B+C$

$K_1=\frac{[B]\times [C]}{[A]}$

b) $B+C\rightarrow D$

$K_2=\frac{[D]}{[B]\times [C]}$

For overall reaction on adding a and b we get c

c) $A\rightarrow D$

$K_3=\frac{[D]}{[A]}$

$K_3=K_1\times K_2=\frac{[B]\times [C]}{[A]}\times \frac{[D]}{[B]\times [C]}$

$K_3=K_1\times K_2=\frac{[D]}{[A]}$

The equilibrium constant for the overall reaction is $K_1\times K_2$

7 0

3 years ago

A material has a density of 8.9 g/cm^3. You have 3 grams of the substance. How much space does it take up?

victus00 [196]

Density (d) is the ratio of the substance mass (m) to its volume (v).
d = m / v
Rearranging the equation gives us the equation for volume which is,
v = m / d
Substituting the values,
v = 3 g / (8.9 g/cm³)
v = 0.34 cm³
The substance occupies approximately 0.34 cm³.

4 0

2 years ago

How to do 3 and 4 questions

Kipish [7]

Answer:

3) Q = -836.8 J.

4) Q = 950J.

Explanation:

Hello there!

In this case, for those calorimetry problems, we use the general equation:

$Q=mC\Delta T$

Thus, we proceed as follows:

3) Here, the temperature difference is from 80 °C to 40 °C, the mass is 5.0 g and the specific heat 4.184 in SI units; thus, the heat is calculated as follows:

$Q=(5g)(4.184\frac{J}{g\°C} )(40\°C-80\°C)\\\\Q=-836.8J$

4) Here, the temperature difference is from 100 °C to 200 °C, the mass is 5.0 g and the specific heat about 1.90 in SI units; thus, the heat is calculated as follows:

$Q=(5g)(1.90\frac{J}{g\°C} )(200\°C-100\°C)\\\\Q=950J$

Regards!

5 0

2 years ago

Assuming a 100% dissociation, what is the freezing point and boiling point of 2.59 m K3PO4(aq).

liubo4ka [24]

Assuming 100% dissociation of K3PO4 you get each molecule of K3 PO4 dissociates into four particles: 3 K4 cations and 1 PO4 - anion. That means that Van't Hoff factor, i, is 4. i = 4. So, the decrease in freezing point of the water solution is Delta Tf = i * Kf * m. Where Kf is the molal cryoscopic constant of water = 1.86 °C /m; and the molality m = 2.59 m => Delta Tf = 4 * 1.86 °C / m * 2.59 m = 19.3 °C. And the freezing point is the normal freezing points less 19.3°C = 0°C - 19.3°C = - 19.3°C. The increase in the boiling point Delta Tb = i * kb * m = 4 * 0.512 °C /m * 2.59 m = 5.3°C => <span>Tb = normal boiling point + 5.3°C = 100°C + 5.3°C = 105.3°C.</span>

3 0

2 years ago

Read 2 more answers