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3 years ago

Ethanol is a:A. substanceB. heterogeneous mixtureC. homogenous mixture

Chemistry

2 answers:

kramer3 years ago

7 0

Answer:

Explanation:

Because ethanol is alcohol and alcohol is a homogenous mixture.

Dahasolnce [82]3 years ago

6 0

Answer: i think it is A

Explanation:

hope this helps!

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A rigid vessel contains water at 0.15 MPa (at 25 °C). The vessel is heated to the critical point of water. Calculate the fractio

pishuonlain [190]

Answer:

0.8078 Kg

Explanation:

Pressure of water = 0.15 MPa = 1.5 bar

At critical point of water ,temperature = 647 K=374°C

From the ideal gas equation

P×V= m×R×T

Let us assume volume = 1 m^3

1.5 x 105 x 1 = m x 287 x 647

m= 0.8078 kg

the fraction of mass of liquid at 25°C.

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3 years ago

A student has a salt water solution. When the student adds more salt to the solution it completely dissolves. What type of solut

AlladinOne [14]

Answer:

<h2>Saturated</h2>

Explanation:

<em>A</em><em>saturated solution is a solution that contains the maximum amount of solute that is capable of being dissolved. Table salt (NaCl) readily dissolves in water.</em>

3 0

2 years ago

Read 2 more answers

Calculate the molarity of a solution of Nach if it contains 7.2.g Nach in 100.0 mL of solution. andver: m Nach . .

Norma-Jean [14]

Answer:

1.23 M

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of NaCl = 7.2 g

As we know , the molecular mass of NaCl = 58.5 g/mol

Moles is calculated as -

n = w / m = 7.2 g / 58.5 g/mol = 0.123 mol

Molarity is calculated as -

V = 100ml = 0.1 L (since , 1 ml = 1/1000L )

M = n / V = 0.123 mol / 0.1 L = 1.23 M

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3 years ago

An archaeologist graduate student found a leg bone of a large animal during the building of a new science building. The bone had

Vlad [161]

Answer : The time passed in years is $2.74\times 10^2\text{ years}$

Explanation :

Half-life of carbon-14 = 5730 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = ?

a = initial amount of the reactant disintegrate = 15.3

a - x = amount left after decay process = 14.8

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{15.3}{14.8}$

$t=274.64\text{ years}=2.74\times 10^2\text{ years}$

Therefore, the time passed in years is $2.74\times 10^2\text{ years}$

4 0

3 years ago

This diagram shows the forces acting on a car. The center dot represents the car, and the arrows represent the forces acting on

Mariulka [41]

Answer : The correct option is B- Towards the left.

Solution : Given,

Force on car in upward direction, $F_1$ = 6000 N

Force on car in downward direction, $F_2$ = 6000 N

Force on car in right direction, $F_3$ = 5000 N

Force on car in left direction, $F_4$ = 6000 N

Force : Force is equal to the mass multiplied by the acceleration. The force is directly proportional to the acceleration.

Formula : F = m × a

where, F is force, m is mass and a is acceleration.

From the given diagram we conclude that the $F_1$ and $F_2$ are in opposite direction and the force acting on car are same. So, their net force is zero.

And in case of $F_3$ and $F_4$ , the net force will not be zero because force acting on car are different. The car will accelerate towards left direction because $F_4$ is higher than the $F_3$ .

So, the car will accelerate towards left direction.

3 0

2 years ago

Ethanol is a:A. substanceB. heterogeneous mixtureC. homogenous mixture​

Ethanol is a:A. substanceB. heterogeneous mixtureC. homogenous mixture