Answer:
See the answer below
Explanation:
<em>The duration of flushing the eyes at the eyewash station in case of accidental contact with chemicals depends of the nature of the chemical.</em>
If the chemical is known to be a <u>non-irritant or mild-irritant</u> one, a <u>5-minute </u>washing time is recommended as the first aid. before seeking the help of a physician,
For <u>moderate to severe irritant</u> chemicals, an immediate <u>15-20 minutes</u> washing period is recommended before seeking further medical help.
For <u>corrosive and strong alkalis</u> chemicals, <u>30 and 60 minutes</u> washing are recommended respectively before seeking the attention of a physician.
However, if the nature of the chemical is unknown,<u> a minimum of 20-minutes washing is generally recommended</u> as first aid before seeking immediate medical help.
1 yard is 91.44 centimeters
Answer:

Explanation:
Volume of a cone:
We have
and we want to find
when the height is 2 cm.
We can see in our equation for the volume of a cone that we have three variables: V, r, and h.
Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.
We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.
Plug this value for r into the volume formula:
Differentiate this equation with respect to time t.
Plug known values into the equation and solve for dh/dt.
Divide both sides by 100π to solve for dh/dt.
The height of the cone is increasing at a rate of 1/10π cm per second.
<h3>
Answer:</h3>
= 5.79 × 10^19 molecules
<h3>
Explanation:</h3>
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
<h3>Step 1: Calculate the number of moles of the compound </h3>

Therefore;
Moles of the compound will be;

= 9.615 × 10⁻5 mole
<h3>Step 2: Calculate the number of molecules present </h3>
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules