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3 years ago

How will adding different flavors of kool-aid to water affect the water's boiling point for kids?

Chemistry

2 answers:

Digiron [165]3 years ago

6 0

Answer:

Kool aid affects the boiling point of water due to its colligative properties.

Explanation:

Colligative properties are the properties of solutions that depend on the number of dispersed particles and are independent of the nature of the solute particles.

Because of the colligative properties, the effects of solute (kool-aid) alter the physical properties of water. An example of this is that kool aid raises the boiling temperature of water. This is because Kool-aid contains citric acid, calcium phosphate, maltodextrin and salt. All of these substances have high boiling points, so when mixed with water they increase it boiling point.

Harlamova29_29 [7]3 years ago

4 0

Adding different flavors of Kool-Aid to water will change boiling point, Kool-Aid increase the water's boiling point because ingredients of Kool-Aid are: citric acid, calcium phosphate, maltodextrin, salt, that substances increase boiling temperature of water.
Kool-Aid is flavored drink mix.

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The work function of an element is the energy required to remove an electron from the surface of the solid element. The work fun

Anton [14]

Explanation:

The given data is as follows.

Energy needed for 1 mole = 279.7 kJ = $279.7 \times \frac{1000 J}{1 kJ}$

= 279700 J

Therefore, energy required for 1 atom will be calculated as follows.

$\frac{279700}{6.022 \times 10^{22}}$

= $4.645 \times 10^{-19} J$

As relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

where, h = planck constant = $6.62 \times 10^{-34} Js$

c = speed of light = $3 \times 10^{8} m/s$

$\lambda$ = wavelength

Therefore, putting given values into the above formula as follows.

E = $\frac{hc}{\lambda}$

$4.645 \times 10^{-19} J$ = $\frac{6.62 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $4.28 \times 10^{-7} m$

or, = $428 \times 10{-9}$ m

= 428 nm

Thus, we can conclude that the maximum wavelength of light that can remove an electron from an atom on the surface of lithium metal is 428 nm.

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3 years ago

The mass of 1.49 of H2 gas at stp

FinnZ [79.3K]

Answer:

Explanation:

I'll assume 1.49 is liters.

All gases occupy 22.4 L for each mole of gas at STP. This makles a very useful, and important, conversion factor:

(22.4L/mole gas) for all gases at STP.

Therefore 1.49L of H2 would be:

(1.49L H2)/(22.4L/mole gas) = 0.0665 moles

The molar mass of H2 is 2, so:

(0.0665 moles)*(2g/mole) = 0.133 grams to 3 sig figs

3 0

3 years ago

At what temperature will the following reaction happen spontaneously, given that ΔH = 8.91×103 J/mol and ΔS = –219.20 J/mol·K? C

Scrat [10]

In order for a certain reaction to be spontaneous, the change in Gibb's free energy must be negative, that is:
dG < 0

We can calculate dG using:
dG = dH - dS

When we plug in the given values, we see that the sign of dG will never be negative; therefore, the reaction will not be spontaneous under any conditions.

4 0

4 years ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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2 years ago

How many moles of N2 are needed to make 5.0 moles of NH3?

Juli2301 [7.4K]

Answer:

The answer is "2.5 mole"

Explanation:

The reaction for producing $NH_3$ can be defined as follows:

Reaction:

$N_2+ 3H_2 \rightarrow 2NH_3$

According to the above reaction, to produce 2 moles of $NH_3$ we need 1 mole of nitrogen:

So, according to the question to produce 5.0 mole $NH_3$ the required $N_2$ :

$\Rightarrow \ 5.0 \ mole \ of \ NH_3 \times \frac{1 \ mole\ of \ N_2}{2 \ mole \ of NH_3}\\\\\Rightarrow 5.0 \times \frac{1}{2}\\\\\Rightarrow 2.5\ mole\\$

To produce 5.0 mole $NH_3$ we need 2.5 mole $N_2$

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3 years ago