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3 years ago

In case of accidental contact with chemicals, how long should the eyes be flushed at the eyewash station

Chemistry

1 answer:

FromTheMoon [43]3 years ago

6 0

Answer:

See the answer below

Explanation:

The duration of flushing the eyes at the eyewash station in case of accidental contact with chemicals depends of the nature of the chemical.

If the chemical is known to be a non-irritant or mild-irritant one, a 5-minute washing time is recommended as the first aid. before seeking the help of a physician,

For moderate to severe irritant chemicals, an immediate 15-20 minutes washing period is recommended before seeking further medical help.

For corrosive and strong alkalis chemicals, 30 and 60 minutes washing are recommended respectively before seeking the attention of a physician.

However, if the nature of the chemical is unknown, a minimum of 20-minutes washing is generally recommended as first aid before seeking immediate medical help.

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0.000431 L Express your answer as an integer.

Molodets [167]

A whole number, not a fraction, that can be negative, positive or zero are integers. They cannot have decimal places.

Now, converting 0.000431 L to decimal an integer as:

$0.000431 L = 431\times 10^{-6} L$

Since, $1 L = 10^{6} \mu L$

So, $431\times 10^{-6} L\times \frac{10^{6 \mu L}}{1 L} = 431 \mu L$ .

Hence, the integer value for 0.000431 L is $431 \mu L[$ .

3 0

3 years ago

How many moles (of molecules or formula units) are in each sample? 79.34 g cf2cl2?

Alona [7]

From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of fluorine = 18.99 grams
molecular mass of chlorine = 35.5 grams
Therefore:
one mole of CF2Cl2 = 12 + 2(18.99) + 2(35.5) = 120.98 grams
Therefore, we can use cross multiplication to find the number of moles in 79.34 grams as follows:
mass = (79.34 x 1) / 120.98 = 0.6558 moles

Now, one mole contains 6.022 x 10^23 molecules, therefore:
number of molecules in 0.65548 moles = 0.6558 x 6.022 x 10^23
= 3.949 x 10^23 molecules

7 0

3 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$