All categories

3 years ago

Select all safety steps below that help you handle chemicals safety

2 answers:

7 0

Answer : For the experiment to study the diffusion across the semi permeable membrane, following are the safety steps that will help to handle chemicals safety;

Use droppers to transfer small amounts of chemicals into another glassware, it will prevent the unnecessary spillage of the chemical.
Use tongs to move hot objects as they may cause severe burns.
Check the test tubes for chips prior to using them otherwise it may leak the chemical inside it
Washing your hands after you finishing the lab experiment, this will ensure that food is not contaminated while consuming, with the chemicals that were stuck on the hands.

Serga [27]3 years ago

4 0

Answer:

select all of them correct

Explanation:

Hope this helps -_-

You might be interested in

Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl

frez [133]

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is $2.29\times 10^{4} gal$ .

Since, 1 gal= 3785.41 mL

Thus, $2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL$

Density of water is 1 g/mL thus, mass of water will be $8.66\times 10^{7}g$ .

Since, 1 grams of chlorine → $10^{6}$ grams of water.

1 g of water → $10^{-6}$ g of chlorine and,

$8.66\times 10^{7}g$ of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

$V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL$

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

6 0

3 years ago

How much heat energy is required to raise the temperature of 0.360 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co

AVprozaik [17]

MThe heat energy required to raise the temperature of 0.36Kg of copper from 22 c to 60 c is calculate using the following formula

MC delta T
m(mass)= 0.360kg in grams = 0.360 x1000 = 360 g
c(specific heat energy) = 0.0920 cal/g.c
delta T = 60- 23 = 37 c

heat energy is therefore= 360g x0.0920 cal/g.c x 37 c= 1225.44 cal

5 0

3 years ago

The vapor pressure of a substance is measured over a range of temperatures. A plot of the natural log of the vapor pressure vers

Gala2k [10]

Answer:

28.7664 kJ /mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,

P is the vapor pressure

ΔHvap is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.

Given :

Slope = -3.46×10³ K

So,

- ΔHvap/ R = -3.46×10³ K

ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol

6 0

3 years ago

Heisenberg's uncertainty principle tells us that _____. heisenberg's uncertainty principle tells us that _____. the de broglie w

Alexandra [31]

Based on Heisenberg's uncertainty principle, the position and velocity of a particle cannot be determined simultaneously with accuracy.

In other words, Heisenberg's uncertainty principle states that the more accurately we know the position of a particle the less accurately we can know its velocity. Mathematically it is given as:

Δx.mΔv >= h/2π

where: Δx = uncertainty in position

m = mass

Δv = uncertainty in velocity

h = plancks constant

3 0

4 years ago

Read 2 more answers

A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k

kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

Helium: 0,25 moles × $\frac{4 g}{1 mol}$ = 1 g of Helium
Argon: 0,25 moles × $\frac{40 g}{1 mol}$ = 10 g of Argon
Krypton: 0,25 moles × $\frac{83,8 g}{1 mol}$ = 20,95 g of krypton
Xenon: 0,25 moles × $\frac{131,3 g}{1 mol}$ = 32,825 g of Xenon