All categories

4 years ago

Why do helium-filled balloons float in air?

2 answers:

8 0

A helium balloon displaces the amount of air. The weight of the helium in the balloon + the balloon fabric = lighter than the air that it displaces. Helium is lighter than air, so therefore, if it's in a balloon, the balloon will float.

kati45 [8]4 years ago

7 0

The helium balloon displaces an amount of air. As long as the weight of the helium plus the balloon fabric is lighter than the air it displaces, the balloon will float in the air. It turns out helium is a lot lighter than air.

You might be interested in

A kno3 solution containing 51 g of kno3 per 100.0 g of water is cooled from 40 ∘c to 0 ∘c. what will happen during cooling?

Gre4nikov [31]

Answer:

As, the temperature decreased from 40.0 °C to 0.0 °C an amount will be recrystallized and precipitated as solid crystals in the water (51.0 g - 14.0 g = 37.0 g) and 14.0 g will be dissolved in water.

Explanation:

Firstly, we must mention that:

The solubility of KNO₃ per 100.0 g of water at 40.0 °C = 63.0 g.

The solubility of KNO₃ per 100.0 g of water at 0.0 °C = 14.0 g.

So, at 40.0 °C, 51.0 g of KNO₃ will be completely dissolved in water.

<em>As, the temperature decreased from 40.0 °C to 0.0 °C an amount will be recrystallized and precipitated as solid crystals in the water (51.0 g - 14.0 g = 37.0 g) and 14.0 g will be dissolved in water.</em>

5 0

3 years ago

Is the movement of plates to form land masses is called plate tectonics

joja [24]

Yes the movement of land forms is called the movment of plate tectonics. :) plz thnk me

4 0

4 years ago

Do respiration in cells releases energy or absorbs energy

igomit [66]

Energy is released during the reaction and is captured by the energy-carrying molecule ATP (adenosine triphosphate).

4 0

4 years ago

In which compound does chlorine have the highest oxidation state?

Temka [501]

NACL0.

Hope this is right! :)

6 0

4 years ago

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

Troyanec [42]

$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the <em>change</em> in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: $0.337\;\text{mol}\cdot\text{dm}^{-3}$ .

7 0

3 years ago