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Ivahew
1 month 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:
castortr0y [3K]1 month ago
6 0
Density is defined as the "mass per unit volume" of an object.

Thus, for an object weighing 100 grams with a volume of 100 milliliters, the density calculates to 100 grams / 100 ml.

When weighing the object, if there is water on the scale's surface, it will contribute additional weight, making the object seem heavier than its actual mass. Consequently, you might mistakenly conclude that the density is GREATER than it truly is.

For instance, if there were 5 ml of water on the scale, with water's density being 1 gram per milliliter (1 g/ml), it would add 5 grams to the object's weight. Using the previous example, the object's mass appears as 105 grams instead of 100 grams. Thus, you would calculate:

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

Thus, the effect on density would be to misleadingly suggest it is greater.

I hope this is helpful!

Best of luck
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When small pieces of manganese oxide were added to a solution of hydrogen peroxide, water and oxygen were produced. Manganese ox
lions [2927]
In the reaction mentioned, increasing the solution's quantity probably increases the reaction rate. This is because a larger number of reactant molecules increases the likelihood of collisions, thereby accelerating the reaction rate. For instance, when manganese oxide is added to hydrogen peroxide, a greater amount of reactants leads to a faster reaction.
4 0
1 month ago
a 0.5678 of KHP required 26.64cm³ of NaOH to complete neutralization.calculate the molarity of the NaOH solution​
lions [2927]

Answer:

Explanation:

0.5678 G        X GRAMS

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 MOL               1 MOL

0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4

0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE  KHC8H4O4=0.00278 MOL NaOH

0.00278 MOL NaOH/26.26ml=0.106 molar

4 0
2 months ago
When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are produced. Determine the enthalpy change per mole o
Alekssandra [3086]

Answer: The change in enthalpy for each mole of zinc involved in the reaction is 152.4 kJ/mol.

Explanation:

First, we need to determine the moles of Zn and HCl.

\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}

The molar mass of Zn is 65 g/mole

\text{Moles of }Zn=\frac{1.34g}{65g/mole}=0.0206mole

and,

\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.750M\times 0.0600=0.0450mole

Next, we must identify the limiting reagent and the excess reagent.

The chemical reaction given is:

Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)

According to the balanced reaction we find that

1 mole of Zn reacts with 2 moles of HCl

Thus, 0.0206 moles of Zn react with 0.0206\times 2=0.0412 moles of HCl

This leads us to determine that HCl is the excess reagent because the moles provided exceed the required moles, while Zn is limiting and restricts product formation.

Now to find the enthalpy change for each mole of zinc reacting in this reaction.

From the reaction we gather that,[ [TAG_59]]

0.0206 moles of Zn yield heat = 3.14 kJ

This implies that 1 mole of Zn generates heat = \frac{3.14kJ}{0.0206mol}=152.4kJ/mol

Hence, the enthalpy change per mole of zinc involved in this reaction amounts to 152.4 kJ/mol.

5 0
2 months ago
Find the specific heat of sikver in J/g C if 38.5 cal is required to heat 25.0g of silver from 31.5 C to 58.7 C (1cal=4.184 J)
alisha [2963]

Answer:

The specific heat value for silver is 0.236 J/g-C.

Explanation:

Silver has a mass of 25 grams.

The temperature shifts from 31.5° C to 58.7° C.

The required heat is equivalent to 25 g.

To determine silver's specific heat, the following equation applies:

Q=mc\Delta T

Where c represents the specific heat of silver.

c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{38.5 \times 4.184}{25(58.7-31.5)}\\\\c=0.236\ J/g-C

Thus, the specific heat of silver is 0.236 J/g-C.

7 0
2 months ago
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