# An object with a mass of 3.5 g raises the level of water in a graduated cylinder from 25.1 mL to 31.7 mL. What is the density of the object? _____g/mL

density of the object = 0.53 g/mL

Explanation:

First we need to find the volume of the object which is equal to the rise of the water in the graduated cylinder:

volume = 31.7 - 25.1 = 6.6 mL

Now we calculate the density of the object using the following formula:

density = mass / volume

density = 3.5 g / 6.6 mL

density = 0.53 g/mL

density

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## Related Questions

What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?

The balanced equation for the above reaction is as follows;
2HCl + K₂SO₃ ---> 2KCl + H₂O + SO₂
stoichiometry of HCl to SO₂ is 2:1
number of moles of HCl reacted - 15.0 g / 36.5 g/mol = 0.411 mol
according to molar ratio
number of SO₂ moles formed - 0.411 mol /2 = 0.206 mol
since we know the number of moles we can find volume using ideal gas law equation
PV = nRT
where
P - pressure - 1.35 atm x 101 325 Pa/atm = 136 789 Pa
V - volume
n - number of moles - 0.206 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation

136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K
V = 4.07 L
volume of SO₂ formed is 4.07 L

Predict the products of each of the following reactions and then balance the chemical equations. (a) Fe is heated in an atmosphere of steam.
(b) NaOH is added to a solution of Fe(NO3)3.
(c) FeSO4 is added to an acidic solution of KMnO4.
(d) Fe is added to a dilute solution of H2SO4.
(e) A solution of Fe(NO3)2 and HNO3 is allowed to stand in air.
(f) FeCO3 is added to a solution of HClO4.
(g) Fe is heated in air.

Answer: The chemical equations are given below.

Explanation:

• For a: Fe is heated in an atmosphere of steam.

The chemical equation follows:

By Stoichiometry of the reaction:

3 moles of iron reacts with 4 moles of steam to produce 1 mole of iron (II,III) oxide and 4 moles of hydrogen.

• For b: NaOH is added to a solution of

The chemical equation follows:

By Stoichiometry of the reaction:

1 mole of iron nitrate reacts with 3 moles of sodium hydroxide to produce 1 mole of iron hydroxide and 3 moles of sodium nitrate.

• For c: is added to an acidic solution of

The chemical equation follows:

By Stoichiometry of the reaction:

10 moles of iron(II) sulfate reacts with 2 moles of potassium permangante and 8 moles of sulfuric acid to produce 5 moles of iron(III) sulfate, 1 mole of potassium sulfate, 2 moles of manganese oxide and 8 moles of water.

• For d: Fe is added to a dilute solution of

The chemical equation follows:

By Stoichiometry of the reaction:

1 mole of iron reacts with 1 mole of sulfuric acid to produce 1 mole of iron sulfate and 1 mole of hydrogen gas.

• For e: A solution of and is allowed to stand in air.

The chemical equation follows:

By Stoichiometry of the reaction:

3 moles of iron(II) nitrate reacts with 4 moles of nitric acid to produce 3 moles of iron(III) nitrate, 1 mole of nitric oxide and 1 mole of water.

• For f: is added to a solution of

The chemical equation follows:

By Stoichiometry of the reaction:

1 mole of iron(II) carbonate reacts with 2 moles of perchloric acid to produce 1 moles of iron(II) perchlorate, 1 mole of carbon dioxide and 1 mole of water.

• For g: Fe is heated in air

The chemical equation follows:

By Stoichiometry of the reaction:

2 moles of iron reacts with 1 mole of oxygen gas to produce 2 moles of iron(II) oxide.

Hence, the chemical equations are given above.

In what stage of wastewater treatment is 90 percent of the oxygen-demanding substances removed from the waste by microbes? primary treatment. secondary treatment. tertiary treatment. septic tank treatment.

I think the correct answer would be secondary treatment. Oftentimes at this stage, microbes are being used to treat the wastewater. It is the heart of the process since a big part of the harmful substances are lowered at this stage. Hope this helps.
I need an answer for this question. here are the choices

primary treatmentsecondary treatmenttertiary treatmentseptic tank treatment

A 20.0g sample of mercury(II) oxide (HgO, M = 216.6) is heated strongly, causing it to decompose to metallic Hg and O2 gas. What volume of O2 gas is produced( measured at STP) (A) 1.03L (B) 2.07L (C) 4.14L (D) 14.0L
When 30.0 ml of 0.10 M AgNO3 is added to 30.0 mL of 0.10 M NaCl, aqueous NaNO3 and solid AgCl are formed. How much solid AgCl is produced?
(A) 0.003 mol (B) 0.0060 (C) 0.030 mol (D) 0.060

For the first reaction, we use it stoichiometric coefficients and the molar mass of zinc. Also, we use the conditions at STP that 1 mol of a substance is equal to 22.4 L. Calculation is as follows:20.0 g (1 mol / 216.6 g)(1 mol O2 / 2 mol Hg)(22.4 L O2 / 1 mol O2) = 1.03 L

For the second reaction, we use also the stiochiometric coefficients to find the amount of AgCl produced from the given conditions.(0.10 mol AgNO3 / 1L soln) (30.0 mL soln) (1L / 1000 mL) ( 1 mol AgCl / 1 mol AgNO3) =0.003 mol AgCl

The oxidation of the sugar glucose, C6H12O6, is described by the following equation. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ΔH = −2802.5 kJ/mol The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body. (a) How much heat in kilojoules can be produced by the metabolism of 18.1 g of glucose?

(b) How many Calories can be produced by the metabolism of 18.1 g of glucose?

(a) 282 kJ

(b) 67.4 Calories

Explanation:

(a) The molar enthalpy, ΔH = −2802.5 kJ/mol, means that the heat produced by the reaction is 2802.5 kJ per mol of glucose.

We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):

(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ

(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.

(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories

For a: The amount of heat produced for given amount of glucose is -283.05 kJ

For b: The amount of heat produced for given amount of glucose is -67648.9 Cal

Explanation:

To calculate the number of moles, we use the equation:

Given mass of glucose = 18.1 g

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

The given chemical reaction follows:

• For a:

By Stoichiometry of the reaction:

When 1 mole of glucose is reacted, the amount of heat released is 2802.5 kJ

So, when 0.101 moles of glucose is reacted, the amount of heat released is

Hence, the amount of heat produced for given amount of glucose is -283.05 kJ

• For b:

To convert the heat produced in kilo joules to calories, we use the conversion factor:

1 kJ = 239 Cal

So,

Hence, the amount of heat produced for given amount of glucose is -67648.9 Cal

when wax is heated, it turns into a liquid. how are the wax molecules affected by this change of state?

Explanation:

Wax is a solid, so it's molecules will be held closer to each other by strong intermolecular forces of attraction.

As a result, when cooled down then wax obtains a define shape and volume.

On the other hand, when wax is heated then its molecules gain kinetic energy due to which there occurs more number of collisions between its particles. As a result, its molecules will move away from each other.

This will lead to change in state of wax from solid to liquid.

Wax molecules have most of their weak van der waals interactions and other hydrophobic interactions decrease and thus most of them are broken between these lipid substances, allowing for the increased molecular movement of these substances. As there are fewer interactions between similar compounds.

Why are the satellites that orbit in the exosphere important

It's the outermost layer of atmosphere, it's thin, but it's said to absorb harmful radiation coming from the Sun, which is really important considering Sun's radiation could melt down the DNA of almost all form of life.

Are The Reactions Of Ordinary Molecular Hydrogen Slow Or Rapid ? Why ?