All categories

3 years ago

If an object is more dense than water, does the object float? 25 point question

Chemistry

1 answer:

LenaWriter [7]3 years ago

5 0

Answer:

<em>If an object is denser than water it will float</em>

Explanation:

You might be interested in

For every action there is an equal and opposite reaction is known as? *

wolverine [178]

Answer:

Newton's third law

3 0

3 years ago

Read 2 more answers

What is the mass of one mole of H

arsen [322]

Answer:

1.008 grams

Explanation:

This numbers are either in grams per mole of atoms or atomic mass unit

4 0

3 years ago

A student gains 1.0 lb in two weeks. How many grams did he gain?

OlgaM077 [116]

1 lb=453.592 grams
So he gained 453.592 pounds

3 0

3 years ago

Read 2 more answers

Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm e

Delicious77 [7]

Answer:

The mass percentage of the solution is 10.46%.

The molality of the solution is 2.5403 mol/kg.

Explanation:

A bottle of wine contains 12.9% ethanol by volume.

This means that in 100 mL of solution 12.9 L of alcohol is present.

Volume of alcohol = v = 12.9 L

Mass of the ethanol = m

Density of the ethanol ,d= $0.789 g/cm^3=0.789 g/mL$

$1 cm^3=1 mL$

$m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g$

Mass of water = M

Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL

Density of water = D=1.00 g/mL

$M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g$

Mass percent

$(w/w)\%=\frac{m}{m+M}\times 100$

$\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%$

Molality :

$m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}$

M = 87.1 g = 0.0871 kg (1 kg =1000 g)

$=\frac{10.1781 g}{46 g/mol\times 0.0871 kg}$

$m=2.5403 mol/kg$

4 0

3 years ago

A silver nitrate, AgNO3, solution of unknown concentration was discovered in the lab. To determine the concentration of the solu

harina [27]

Answer:

0.174 M

Explanation:

The same can be solve by using Nernst's equation

The Nernst's equation is:

$Ecell=E^{0}_{cell}-\frac{0.0592}{n}log\frac{[anodic]}{[cathodic]}$

For silver cell

n= 1

As both the compartments have silver nitrate solution,t the standard emf of cell will be zero.

Given

Ecell = 0.045

[Anodic compartment]= 1 M

Putting values

$0.045 = 0 -\frac{0.0592}{1}log(\frac{1}{x} )$

$0.760=log(\frac{1}{x})$

Taking antilog and solving

[AgNO3]=0.174 M

3 0

4 years ago