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sergiy2304 [10]
3 years ago
11

What instruments can be used to measure volumes of liquids?

Chemistry
2 answers:
TiliK225 [7]3 years ago
8 0
Graduated cylinders or beakers. 
alexandr1967 [171]3 years ago
5 0
Graduated Cylinders, Measuring Cups, Measuring Spoons, Beakers, Flasks
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The pulpy acidic fluid which passes from the stomach to the small intestine, consisting of gastric juices and partly digested food.
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Which is an element hydrogen , air , carbon dioxide , water​
Basile [38]

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Hydrogen is an element

Explanation:

Hydrogen is an element with only hydrogen atoms, whereas air, carbon dioxide, and water are all made up of multiple elements with different types of atoms.

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5 points<br> What is the name of the atom pictured here? Use a periodic table if<br> needed. *
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Explanation:

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3 years ago
What is the number of grams of an element that is numerically equal to the atomic mass in amu?
bagirrra123 [75]
The wording of your question doesn't quite make sense, but a mole of an element has the same mass in grams as a single atom of that element has in amu. The mole is defined as 6.02 x10^22 things, whether they be atoms or molecules or even moles! 6.02x10^22 atoms of carbon has a mass of 12.01 g, and a single atom of carbon has a mass of 12.01 amu. Hope this helps!
3 0
3 years ago
Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2
baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) Cu_{(s)} →Cu^{2+}_{(aq)} +2e E°=0.337 v

(reduction) MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}→MnO_{2 (aq)}+2H_{2}O E°=1.679 v

(overall) 2MnO_{4 (aq)}+3Cu_{(s)}+8H^{+}_{(aq)}→3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346

E=1.346

5 0
3 years ago
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