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LuckyWell [14K]
2 years ago
7

There is water on the pan of the scale as you measure the mass of an object. If you were to ignore the water, what would be the

effect on your density calculation?
Chemistry
1 answer:
mihalych1998 [28]2 years ago
6 0
Remember that density refers to the "mass per unit volume" of an object.

So, if an object had a mass of 100 grams and a volume of 100 milliliters, the density would be 100 grams / 100 ml.

In the question, water on the surface of the scale would add weight, so the mass of the object that you're weighing would appear to be heavier than it really is. If that happens, you'll incorrectly assume that the density is GREATER than it really is

As an example, suppose that there was 5 ml of water on the surface of the scale. Water has a density of 1 gram per milliliter (1 g/ml) so the water would add 5 grams to the object's weight. If we use the example above, the mass of the object would seem to be 105 grams, rather than 100 grams. So, you would calculate:

density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml

The effect on density would be that it would erroneously appear to be greater

Hope this helps!

Good luck
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You are given a compound with the formula mcl2, in which m is a metal. you are told that the metal ion has 26 electrons. what is
Nookie1986 [14]

Answer:

             Nickle

Explanation:

                  In statement it is given that the metal "M" has a formula "MCl₂" which means it has a charge of +2 as each Chlorine atom has -1 charge. Therefore, it is cleared that this metal has lost 2 electrons and originally it was carrying 28 electrons.

                  After that periodic table is concerned and a metal with 28 atomic number was tracked because a neutral atom with 28 electrons must have 28 protons and protons are infact the atomic number.

                  Hence, in periodic table the metal with 28 atomic number is found to be Nickle.

6 0
2 years ago
Which of the reagents listed below would efficiently accomplish the transformation of ethyl-3-pentenoate into 3-penten-1-ol?
Mandarinka [93]
Reactions of Ethyl-3-pentenoate with all given reagents are given below.

Reaction with H₂ / Pd:
                                     
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.

Reaction with NaBH₄:
                                   Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.

Reaction with LiAlH₄:
                                  Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.

Result:
           The correct answer is Option-A (Highlighted RED below).

7 0
2 years ago
What is the oxidation state of selenium in SeO3?​
ra1l [238]

Answer: The oxidation state of selenium in SeO3 is +6

Explanation:

SeO3 is the chemical formula for selenium trioxide.

- The oxidation state of SeO3 = 0 (since it is stable and with no charge)

- the oxidation number of oxygen (O) IN SeO3 is -2

- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)

Hence, SeO3 = 0

Z + (-2 x 3) = 0

Z + (-6) = 0

Z - 6 = 0

Z = 0 + 6

Z = +6

Thus, the oxidation state of selenium in SeO3 is +6

8 0
2 years ago
Catherine has some seawater. She wants to separate pure water from the seawater.
Flura [38]

Answer:

the change is evaporation

Explanation:

the water heats up at the surface of the water and evaporates

4 0
2 years ago
Read 2 more answers
A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a vo
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Answer:

58.6 % by mass of Na₂CO₃

Explanation:

This is the reaction:

Na₂CO₃  +  MgCO₃ +  4HCl  →  MgCl₂  +  2NaCl  + 2CO₂  +  2H₂O

Let's find out the moles of CO₂ produced, by the Ideal Gases Law

1.24 atm . 1.67 L = n . 0.082 . 299K

(1.24 atm . 1.67 L / 0.082 . 299K) = n

0.0844 moles = n

Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:

2 moles of CO₂ were produced by 1 mol of Na₂CO₃

Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2  = 0.0422 moles of Na₂CO₃.

Let's convert this moles into mass (mol . molar mass)

0.0422 mol . 106 g/mol = 4.47 g

Finally we can know the mass percent of sodium carbonate in the mixture

(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %

3 0
1 year ago
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