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3 years ago

Which has the lowest freezing point?

1 answer:

8 0

Freezing point depression depends of the number of particles of the solute in the solution.
1)Pure water have highest freezing point. All other solutions with given solutes will have lower temperatures.
2) The more particles of the solute in the solution the lower freezing point is going to be.
b. 1.0 m NaCl ( dissociates and give 2 mol ions (1 mol Na⁺ and 1 mol Cl⁻))
c. 1.0 m K3PO4 (dissociates and give 4 mol ions (3 mol K⁺ and 1 mol PO4³⁻)
d. 1.0 m CaCl2 (dissociates and give 3 mol ions (1 mol Ca²⁺ and 2 mol Cl⁻))
e. 1.0 m glucose (c6h12o6) (glucose does not dissociate, and solution have
1 mole of particles of the solute(glucose))

The largest number of particles has 1.0 m K3PO4 solution, and it is has lowest freezing point . Answer is C.

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You have used 2.4×102 L of distilled water for a dialysis patient. How many gallons of water is that?

Alinara [238K]

Answer:

63.36gallons

Explanation:

Given:

Volume of water used for dialysis = 2.4 x 10²L

Solution:

We are to convert from liters to gallons.

The conversion factor is shown below:

1L = 0.264gallons

To convert to litre:

since 1L = 0.264gallons

2.4 x 10²L = (2.4 x 10² x 0.264)gallons; 63.36gallons

3 0

3 years ago

What is the mass in grams of 1.204x10^24 atoms of carbon

Mekhanik [1.2K]

It should be 24g of carbon

4 0

2 years ago

How many moles of N2 you have in 476 L sample at STP?

aleksandrvk [35]

You have to take the 476 L and divide it by 22.4 because with Molar Volume 1 mol = 22.4L. After you divide it your answer is 21.25.

5 0

3 years ago

Which of the following atoms or ions does not have a filled outer subshell?CaSZn2+S2Ca2+

ipn [44]

S and S²⁻ do not have the outer subshell fully filled with electrons.

Explanation:

We look at electronic configurations:

Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons

S 1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons

Zn²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons

S²⁻ 1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons

Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ - the outer subshell 3p⁶ is fully-filled with electrons

Learn more about:

electron configurations

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4 0

3 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

8 0

2 years ago