Answer:
<h3><em>to Separate camphor from sand we use <u>sublimation</u> process.</em></h3><h3 /><h3 /><h3><em>Hope </em><em>it</em><em> </em><em>is </em><em>helpful</em><em> to</em><em> you</em><em> </em></h3>
Answer:
The person should not be concerned about radon.
Explanation:
<em>A person living on the sixth floor of an aparment probably should not be concerned about radon</em>. In the conditions of the Earth's atmosphere (temperature and pressure), radon exists as a gas. This gas has a density that is approximately 8 times higher than the density of air (9.73 g/L compared to 1.22 g/L). <em>This means that radon gas would not rise, and instead remain close to the ground</em>, meaning that an apartment on a sixth floor is too far away from the ground for radon gas to reach there.
A gas being denser than air is also the reason why if you blow into a balloon, it will fall to the ground, because CO₂ is denser than air.
Answer:
A planet's <u>hydrosphere</u> can be<u> liquid</u>, <u>vapor</u>, or <u>ice</u>. On Earth, in the places at the <u>north and south pole</u>, water exists in ice or glacier form, in the <u>atmosphere</u> it exists in vapor form and liquid water exists on the <u>surface</u> in the form of oceans, lakes and rivers. It also exists below ground as <u>groundwater</u>, in wells and aquifers. Water collects in clouds, then falls to Earth in the form of <u>rain or snow</u>,
<em>Hope it helps</em>
<em>:D</em>
<em />
Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
Answer:
0.1313 g.
Explanation:
- It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
- Suppose that hydrogen behaves ideally and at STP conditions.
<u><em>Using cross multiplication:</em></u>
1.0 mol of hydrogen occupies → 22.4 L.
??? mol of hydrogen occupies → 1.47 L.
∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.
- Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:
<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>