All categories

2 years ago

A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.314 kg of water?

Chemistry

1 answer:

Mariulka [41]2 years ago

5 0

Answer:

g NaCl = 424.623 g

Explanation:

<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent

∴ solvent: H2O

∴ mass H2O = 2.314 Kg

mol NaCl:

⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl

∴ mm NaCl = 58.44 g/mol

⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl

You might be interested in

When 50 ml (50 g) of 1.00 m hcl at 22oc is added to 50 ml (50 g) of 1.00 m naoh at 22oc in a coffee cup calorimeter, the tempera

vitfil [10]

Answer:

$\boxed{\text{2700 J}}$

Explanation:

HCl + NaOH ⟶ NaCl + H₂O

There are two energy flows in this reaction.

Heat of reaction + heat to warm water = 0

q₁ + q₂ = 0

q₁ + mCΔT = 0

Data:

m(HCl) = 50 g

m(NaOH) = 50 g

T₁ = 22 °C

T₂ = 28.87 °C

C = 4.18 J·°C⁻¹g⁻¹

Calculations:

m = 50 + 50 = 100 g

ΔT = 28.87 – 22 = 6.9 °C

q₂ = 100 × 4.18 × 6.9 = 2900 J

q₁ + 2900 = 0

q₁ = -2900 J

The negative sign tells us that the reaction produced heat.

The reaction produced $\boxed{\textbf{2900 J}}$ .

7 0

2 years ago

When monomers combine to form a condensation polymer, another product is also formed. typically, this product is _____. methanol

shtirl [24]

Water is the answer!

4 0

2 years ago

Read 2 more answers

What should be done if particles of precipitate appear in the filtrate?

Volgvan

<span>The instructor should be questioned to see if the filtrate is able to be recycled. This precipitate can contaminate the filtrate, rendering it useless for repeated experiments. If it is able to be recycled, a second pass through the filter might be required to remove the precipitate.</span>

4 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

SVETLANKA909090 [29]

Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

3 0

2 years ago