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professor190 [17]

3 years ago

9

Litmus paper is made from water-soluble dyes which are extracted from lichens. This paper is used as an acid-base indicator. Whi

ch of these common household substances would turn blue litmus paper red? A) bleach B) lye C) soap D) vinegar

1 answer:

Nutka1998 [239]3 years ago

3 0

The household substances that will turn blue litmus paper red are: A) bleach and D) Vinegar. This is because bleach contains acidic solutions; and vinegar contains a percentage of organic acids for example, acetic acid.

You might be interested in

Write a paragraph on: how Westfield how the rust in their water formed.

Mariana [72]

Answer:

Students have investigated what's going on in the fictional town of Westfield, they learned that the mysterious reddish-brown substance in the water is actually rust, which formed because of a chemical reaction between the iron pipes and the fertilizer substance in the water.

Explanation:

8 0

3 years ago

A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0

4 years ago

Kepler and Newton studied the movement of planets. Kepler studied the speed of the planets while Newton studied the forces that

nasty-shy [4]

Kepler did not study the speed of the planets, rather, he studied how the planets move in the solar system. He proposed three laws. As a summary, he described that the planets move around the sun in the shape of an ellipse (orbit), and the Sun being one of the foci. Then, he proposed the period for the planet to complete one revolution around the Sun.

On the other hand, Newton studied the forces acting on the planet (or any object in space) that explain how the planets move around the solar system as described by Kepler. Also, Kepler's observations only apply to planets and not the moons or satellites. Thus, Kepler only made laws from observations, while Newton based it from underlying principles that led him to mathematical equations such as the law of universal gravitation.

4 0

3 years ago

Read 2 more answers

Calculate the concentration of CO2 in water at 25 degrees Celsius when the pressure of CO2 over the solution is 4.5 atm. At 25 d

OLEGan [10]

Answer : The concentration of $CO_2$ is, 0.12 M

Explanation :

Using Henry's law :

$C_{CO_2}=k_H\times p_{CO_2}$

where,

$C_{CO_2}$ = concentration of $CO_2$ = ?

$p_{CO_2}$ = partial pressure of $CO_2$ = 4.5 atm

$k_H$ = Henry's law constant = $3.1\times 10^{-2}mol/L.atm$

Now put all the given values in the above formula, we get:

$C_{CO_2}=3.1\times 10^{-2}mol/L.atm\times (4.5atm)$

$C_{CO_2}=0.1395M\approx 0.12M$

Thus, the concentration of $CO_2$ is, 0.12 M

5 0

3 years ago

The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma

Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= $p_o= 187.54 mmHg$

Vapor pressure of the solution at 65 °C= $p_s$

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

$p_s=173.83 mmHg$

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0

3 years ago