All categories

3 years ago

In what situation do you use a pipette

1 answer:

5 0

To answer your question, you use a pipette when controlling how much liquid you use in an experiment. {i.e. trying to put hydrogen peroxide into a glass full of vinegar.}

I don't know if that will cause a chemical reaction, but I hope this answer helps!

You might be interested in

What volume would a sample of gas occupy in LITERS at 0.985 atmospheres and a volume of 3.65 liters if the pressure were raised

musickatia [10]

Answer:

3.18 L

Explanation:

Step 1: Given data

Initial pressure (P₁): 0.985 atm

Initial volume (V₁): 3.65 L

Final pressure (P₂): 861.0 mmHg

Final volume (V₂): ?

Step 2: Convert P₁ to mmHg

We will use the conversion factor 1 atm = 760 mmHg.

0.985 atm × 760 mmHg/1 atm = 749 mmHg

Step 3: Calculate the final volume of the gas

Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L

7 0

3 years ago

What is the charge for MnO4

lakkis [162]

Answer:

Mn = 2+, O = -2

Explanation:

3 0

3 years ago

Describe to me the difference on an atomic/particle level between a Solid and a Gas.

Nitella [24]

Particles in a: gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.!

7 0

3 years ago

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volum

zhuklara [117]

Answer:

The volume of nitrogen oxide formed is 35.6L

Explanation:

The reaction of nitric acid with copper is:

Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)

Moles of copper are:

$4.95cm^3\frac{8.95g}{1cm^3} \frac{1mol}{63.55g} = 0.697 moles$

Moles of nitric acid are:

$230mL\frac{1.42g}{mL} \frac{68g}{100g} \frac{1mol}{63.01g}=3.52moles$

As 1 mol of Cu reacts with 4 moles of HNO₃:

0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.

Moles of NO₂ produced are:

0.697 mol Cu × (2mol NO₂ / 1mol Cu) = 1.394 moles of NO₂

Using PV = nRT

Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K).

Thus, volume is:

V = nRT / P

V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm

V = 35.6L

The volume of nitrogen oxide formed is 35.6L

3 0

3 years ago

A biofuel researcher is running an experiment to produce methane from wastewater treatment plant sludge. The researcher places t

vodomira [7]

Answer: Missing data related to question is attached below

If the KH,CH4 = 1.42 * 10^-3 mol/L atm

answer:

0.88 atm

Explanation:

P1 = 20 psi

aqueous concentration of methane in closed bottle = 20 ppm

20 ppm = 20 / 1000 ( gram / Kg ) = 0.020 gram/kg = 0.020 g/liter

molar mass of methane = 16 gram/mol

next : convert mass conc to molar conc

Cm = 0.020 / 16

Cm = 1.25 * 10^-3 mol/L

Given that KH,CH4 = 1.42 * 10-3 mol/L atm

Applying equilibrium relationship

Cm = ( KH,CH4 ) ( partial pressure of methane )

hence partial pressure of methane

= Cm / ( KH,CH4 ) = 1.25 * 10^3 / (1.42 * 10^-3 ) = 0.88 atm

7 0

3 years ago