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Ira Lisetskai [31]
3 years ago
11

Which of the following would make a solution?

Chemistry
2 answers:
gizmo_the_mogwai [7]3 years ago
8 0

Answer:

B sugar in water

Explanation:

because sugar dissolves in water while the others don't

Oliga [24]3 years ago
5 0
B. Sugar in water
Sugar dissolves in water
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Please help me this is a quiz gde
wel

An not sure,, but maybe 24. don't take my word for it

5 0
2 years ago
Se construye una pila galvánica con una barra de cobre sumergida en una disolución 1M de cationes Fe+2 y una barra de plata sume
pashok25 [27]

Answer:

El potencial celular estándar, E_{cell} is +0.46 V

Explanation:

Las reacciones de media célula son;

Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V

Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V

Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V

y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V

 que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo

Por lo tanto, en el ánodo, tendremos

Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)

En el cátodo

2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)

E_{cell} = E_c + E_a = -0.34 + 0.8 = +0.46 \, V

El potencial celular estándar, E_{cell} = +0.46 V

4 0
3 years ago
Think about the lab procedure you just read. Label each factor below with V for “variable” or C for “constant”.
luda_lava [24]
<h2>Answer:</h2><h3>The temperature of the gas: V</h3>

The temperature of gas is a variable quantity. It can be changed by changing energy or pressure of gas.

<h3>The amount of gas in the tube (in terms of mass and moles): C</h3>

It is a constant entity. As mass of gas once taken can not be changed by changing temperature, pressure etc.

<h3>The radius of the tube: C</h3>

The radius of tube cannot change at any rate.

<h3>The temperature of the gas (changed by the water surrounding it):  V</h3>

It can be changed by changing the temperature of water surrounding it.

<h3>The type of gas: C</h3>

It can never be changed.

<h3>The pressure of the gas: V</h3>

It can be changed by simply changing temperature and volume of gas.

8 0
3 years ago
Read 2 more answers
PLEASE HURRY AND SHOW WORK
IgorC [24]

Answer:

The answer to your question is   P = 1.64 atm

Explanation:

Data

Volume = 2.5 x 10⁷ L

Temperature = 22°C

Pressure = ?

Moles = 1.7 x 10⁶

R = 0.082 atm L/ mol°K

Process

1.- Convert temperature to °K

Temperature = 22 + 273

                      = 295°K

2.- Use the Ideal gas law to solve this problem

                PV = nRT

- Solve for P

                P = nRT / V

- Substitution

                P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷

- Simplification

                P = 41123000 / 2.5 x 10⁷

- Result

                P = 1.64 atm

3 0
3 years ago
2. When heated above 500 ºC, potassium nitrate decomposes according to the equation below. 4KNO3 2K2O + 2N2 + 5O2A.If oxygen is
AysviL [449]

Answer:

(a) The rate of formation of K2O is 0.12 M/s.

The rate of formation of N2 is also 0.12 M/s

(b) The rate of decomposition of KNO3 is 0.24 M/s

Explanation:

(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.

Rate of formation of O2 is 0.3 M/s

Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s

Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.

Rate of formation of N2 = (2×0.3/5) = 0.12 M/s

(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.

Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s

3 0
3 years ago
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