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Dafna1 [17]
2 years ago
5

Consider a mixture of soil and water and compare it to a colloid, such as milk. Which property best differentiates these two mix

tures?
Chemistry
1 answer:
mezya [45]2 years ago
7 0
There are many differentiates between them. Like if you standing these two mixture, the soil and water will separate while milk will not. And colloid have tyndall effect while soil and water doesn't have.
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Why are space probes and other unmanned crafts more commonly used for space investigations than space crafts with crews of astro
Crank
All of these are correct.
4 0
2 years ago
Which of the following is an element?
Kryger [21]

Oxygen is a chemical element with symbol O and atomic number 8. Classified as a nonmetal, Oxygen is a gas at room temperature.

8 0
2 years ago
Answer the following questions that relate to electrochemical reactions. (a) Under standard conditions at 25˚C, Zn(s) reacts wit
Elanso [62]

Answer:

0.48 V

Explanation:

Zn(s) ------------> Zn^2+(aq) + 2e. Oxidation half equation (-0.76V)

Co^2+(aq) + 2e-----------> Co(s). Reduction half equation (-0.28)

Zn(s) + Co^2+(aq) -------------> Zn^2+(aq) + Co(s) overall redox equation

Zinc is the anode while cobalt is the cathode.

E°cell= E°cathode - E°anode

E°cell= -0.28-(-0.76)= 0.48 V

7 0
3 years ago
Read 2 more answers
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

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

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
2 years ago
Convert 1.45 × 1024 atoms of carbon to moles of carbon.
anygoal [31]
1 mol = 6.02 * 10^23 atoms of carbon
x mol = 1.45 * 10^24 atoms of carbon

1/x =6.02*10^23 / 1.45 * 10^24

6.02 * 10^23 x = 1.45 * 10^24
x = 1.45 * 10^24 / 6.02 * 10^23
x = 2.41 mols of carbon
8 0
3 years ago
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