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2 years ago

Assume that there is a puddle that contains 12.0 kg of water and it froze outside since it is -1°C. The

Chemistry

1 answer:

Musya8 [376]2 years ago

4 0

The temperature that must be to freeze the solution would be -21.1 ° C.

<h3>How to calculate the freezing temperature of this solution?</h3>

To calculate the freezing temperature we must take into account the following information.

Solution with a salt concentration of 10% is frozen at -6°C
Solution with a salt concentration of 20% is frozen at -16°C
Solution with a higher concentration is frozen at -21.1°C

According to the above, it can be inferred that the puddle has a 50% concentration of salt because they had 12 kg of water and 6 kg of salt.

So the lowest freezing temperature would be 21.1°C because the puddle is 50% concentrated.

Note: This question is incomplete because there is some missing information. Here is the missing information:

A 10% salt solution freezes at about 20°F (-6°C), and a 20% solution freezes at 2°F (-16°C).
The lowest freezing point obtainable for salt solutions is −21.1 °C

Learn more about freezing in: brainly.com/question/14131507

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Molecules can be either positively or negatively charged depending upon their elemental arrangement

Nuetrik [128]

They are positively or negatively charged based on their electrical configuration of electrons*
For example an electronic configuration of 2,8,3 would have a negative charge if +3 since it needs to lose 3 electrons to gain the electrical configuration of a noble gas
2,8,1 would have a charge of +1 for the same reason
2,8,6 would be -2 since it is easier to gain 2 electrons that lose 6 electrons

Hope this helped :))

8 0

2 years ago

A projectile is fired with speed v0 at an angle theta from the horizontal from the horizontal as shown in the figure.

irga5000 [103]

Answer:

v₀ = √(2gH/(sin²θ)) = (sin θ)√(2gH)

v₀ = √(gR/(sin2θ))

Explanation:

An image of the artillery officer, the hill and path of motionof the projectile is attached to this solution.

Given, R, H, g and θ (theta)

Using the equations of motion, we can get the initial velocity v₀

First of, we need to resolve this motion into the vertical and horizontal axis.

The horizontal component of the initial velocity, v₀ₓ = v₀ cos θ

Vertical component of the initial velocity, v₀ᵧ = v₀ sin θ

When the projectile reaches maximum height, Velocity at max height, vₕ = 0m/s

From equations of motion,

vₕ = v₀ᵧ - gt

0 = v₀ sinθ - gt

t = v₀ sinθ/g

This is the time taken to reach maximum height. The time take to comolete the toyal flight, T = 2t = (2v₀ sinθ)/g

The maximum height to be reached, H can be calculated from the equations of motion too

H = vₕt - 0.5gt² = 0 - 0.5g((v₀ sinθ)/g)²

H = (0.5g v₀² sin²θ)/g²

H = (v₀² sin²θ)/2g

The range, or horizontal distance to be covered by the projectile, R, will be calculated using the horizontal component of the initial Velocity, v₀ₓ = v₀ cos θ, this horizontal velocity is constant all through the motion, so, no acceleration in the horizontal direction.

R = v₀ₓT = (v₀ cos θ)((2v₀ sinθ)/g)

R = (v₀²(2cosθsinθ)/g)

2cosθsinθ = sin2θ

R = v₀²(sin2θ)/g

So, writing v₀ in terms of all the other parameters,

v₀ = √(2gH/(sin²θ)) = (sinθ)√(2gH

v₀ = √(gR/(sin2θ))

4 0

2 years ago

Which of the following is an anion?<br> A. O^2- <br> C. Al^3+ <br> D. H

kifflom [539]

The correct answer is A. O^2-

6 0

2 years ago

a sample of ammonia contains 9g hydrogen and 42g nitrogen. another sample contains 5g hydrogen .calculate the amount of nitrogen

balandron [24]

9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen

$9x=42 \cdot 5 \\
9x=210 \\
x \approx 23.33$

The mass of nitrogen in the second sample is 23.33 g.

4 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0

3 years ago