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Marianna [84]
2 years ago
15

. Divide 94.20 g by 3.167 22 mL.

Chemistry
2 answers:
jeka942 years ago
7 0
94.20 g/3.16722 mL = 29.74 g/mL

The ratio of mass to volume is equal to the substance's density. Thus, 29.74 g/mL is the density of whatever substance it may be. Density does not change for incompressible matter like solid and some liquids. Although, it may be temperature dependent.
sashaice [31]2 years ago
7 0

Answer : The answer will be 29.74 g/mL

Explanation :

As we are given an expression :

Divide 94.20 g by 3.16722 mL.

To solve this problem, we use division operation. So,

\frac{94.20g}{3.16722mL}

By solving the term, we get:

\Rightarrow 29.74g/mL

According to the significant rule of division, the least number of significant figures in any number of the problem determines the number of significant figures in the answer.

In the given expression, 94.20 has 4 significant figures and 3.16722 has 6 significant figures. From this we conclude that 4 is the least significant figures in this problem. So, the answer should be in 4 significant figures.

Thus, the answer will be 29.74 g/mL

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How many molecules are in 3.6 grams of NaCl?
raketka [301]

Answer:

\boxed{\text{None}}

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}

Step 2. Convert moles to formula units

\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}

There are \boxed{3.4 \times 10^{22} \text{ FU NaCl}} in 3.6 g of NaCl.

6 0
2 years ago
4. a student is shown a Diagram which shows gas bubbles being made by water plant that has been exposed to light. which gas is m
lisov135 [29]
Carbon dioxide is in the bubbles
5 0
2 years ago
The density of ice is 0.917 g/cm3. How much volume does 52.3 g of ice occupy? Show your work.
mash [69]

Answer:

The answer is

<h2>57.0 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}

From the question

mass of ice = 52.3 g

density = 0.917 g/cm³

The volume is

volume =  \frac{52.3}{0.917}  \\  = 57.0338058

We have the final answer as

<h3>57.0 mL</h3>

Hope this helps you

6 0
2 years ago
The van der waals constants a and b for benzene are 18.00 atm l2 mol−2 and 0.115 l mol−1, respectively. Calculate the critical c
Aneli [31]

There are three critical constants namely, critical temperature, critical pressure and critical volume.

Critical temperature is defined as temperature of gas below which the increase in pressure cause liquefaction of gas and above that liquefaction of gas do not take place.

Critical pressure is defined as pressure needed to liquefy a gas at critical temperature. Volume of 1 mol of gas at critical pressure and temperature is known as critical volume.

Critical temperature can be calculated as follows:

T_{C}=\frac{8a}{27Rb}

Putting the value,

T_{C}=\frac{8(18.00atm L^{2}mol^{-2})}{27(0.0821 atm L K^{-1} mol^{-1})(0.115 L mol^{-1})}=564.88 K

Thus, critical temperature is 564.88 K.

Critical pressure is calculated as follows:

P_{C}=\frac{a}{27b^{2}}

Putting the values,

P_{C}=\frac{18.00 atm L^{2}mol^{-2}}{27(0.115 L mol^{-1})^{2}}=50.41 atm

Therefore, critical pressure is 50.41 atm.

Now, calculate critical volume as follows:

V_{C}=3b

Putting the values,

V_{C}=3(0.115 L mol^{-1})=0.345 L mol^{-1}

Therefore, for 1 mol critical volume is 0.345 L.

5 0
2 years ago
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kvasek [131]

Answer:

Detail is given below.

Explanation:

Chemical equation:

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This reaction shows that when ammonia react with oxygen it form water and nitrogen gas.

There are two reactants on left hand side oxygen and ammonia. Ammonia is formed when nitrogen and hydrogen react. While on right hand side there are two products nitrogen and water. Water is formed by the reaction of hydrogen and oxygen.

The given reaction also shows that it follow the law of conservation of mass.

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.  

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

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