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2 years ago

Calculate each of the following quantities:

Chemistry

1 answer:

ICE Princess25 [194]2 years ago

8 0

$2.54$ × $10^{24}$ sodium ions in 2.11 mol of sodium carbonate.

<h3>What is sodium carbonate?</h3>

The sodium carbonate, the inorganic substance with the $Na_2CO_3$ formula, is also known as washing soda, soda ash, and soda crystals.

All of the forms are water-soluble, odourless, white salts that produce slightly alkaline solutions in water. In the past, it was recovered from the ashes of plants that were raised in soils high in salt.

The term "soda ash" originated to refer to sodium carbonate because the ashes of these sodium-rich plants were probably distinct from the wood ashes that were originally used to make potash. The Solvay technique is used to produce it in large amounts from limestone and sodium chloride.

Learn more about sodium carbonate $Na_2CO_3$

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In the 1700's, a French chemist named Antoine Lavoisier

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The Law of Conservation of Mass

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3 years ago

A student investigated the diffusion of ammonia gas, NH3, and hydrogen chloride gas, HCl. The damp red litmus paper for NH3 chan

Alex Ar [27]

Answer:

43.96secs

Explanation:

M1 = molar mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

t1 = time for NH3 to diffuse = 30secs

M2 = molar mass of HCl = 1 + 35.5 = 36.5g/mol

t2 = time for HCl to diffuse =?

From Graham's law of diffusion:

t2/t1 = √(M2/M1)

t2/30 = √(36.5/17)

t2/30 = 1.465

Cross multiply

t2 = 30 x 1.465

t2 = 43.96secs

Therefore it will take 43.96secs for the damp blue litmus paper to change colour for HCl.

8 0

3 years ago

A liquid that occupies a volume of 0.820 L has a mass of 2.56 kg.

user100 [1]

Answer:

The density of this liquid is 0.320 kg/L

Explanation:

Given:

Volume of the Liquid = 0.820 L

Mass of the liquid = 2.56 kg.

To Find:

The density of the liquid in kg/L

Solution:

Density is the mass occupied by the substance in unit volume. This density is essential determining whether the substance floats or sinks. Greek letter(rho) is used to denote density

The equation of for density is

$Density = \frac{mass}{volume}$

$\rho = \frac{m}{v}$

where m is the mass

v is the volume

On substituting the given values

$\rho = \frac{0.820 }{2.56}$

$\rho = 0.320 kg/L$

7 0

3 years ago

Two 10g blocks, one of copper and one of iron, were heated from 300 K to 400K (a temperature difference of 100 K).

12345 [234]

1) 385 J

2) 450 J

Explanation:

The amount of energy that must be absorbed by a certain substance in order to increase its temperature by $\Delta T$ is given by the equation:

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is the increase in temperature of the substance

For the block of copper in this problem, we have:

m = 10 g is the mass

$C=0.385 J/gK$ is the specific heat capacity of copper

$\Delta T=400-300 = 100 K$ is the change in temperature

So, the energy absorbed by the block of copper is

$Q=(10)(0.385)(100)=385 J$

Similarly for the block of iron, the energy absorbed by the iron is given by

$Q=mC\Delta T$

where

m is the mass of the block of iron

C is its specific heat capacity of iron

$\Delta T$ is the increase in temperature of the block

Here we have:

m = 10 g is the mass of the block

$C=0.450 J/gK$ is the specific heat capacity of iron

$\Delta T=400-300 = 100 K$ is the change in temperature

So, the energy absorbed by the block of iron is

$Q=(10)(0.450)(100)=450 J$

6 0

3 years ago

List three criteria used to identify a cold front.

zaharov [31]

Answer:

Explanation:

A cold front is a transition zone from warm air masses to cold air masses. We use weather elements to identify cold fronts.

Temperature: Since air moves from warm masses to cold masses, the air in a cold front is at different temperatures. The air behind a cold front is warm while the one ahead is cold. This implies that within a cold front is at different temperatures.
Precipitation: Prior to the passing of cold front, precipitation is usually rife with low showers. A coldfront in itself is usually accompanied by heavy rainfall full of thunderstorms and lightening. After a coldfront, the showers steadies and decreases.
Pressure changes: before a coldfront, the atmospheric pressure decreases steadily. When the front arrives, the pressure further lowers with a sharp increase thereafter. After the front, the pressure can continue to increase.

3 0

3 years ago