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ehidna [41]
2 years ago
14

Calculate how many moles of carbon dioxide are contained in 5.57 x 10²¹ molecules of CO₂

Chemistry
1 answer:
Alex777 [14]2 years ago
3 0
<h2>Answer:  0.00925 mol </h2>

Explanation:

Moles = # of molecules ÷ Avogadro's Number  

         =  (5.57 × 10²¹ molecules) ÷ (6.022 × 10²³ molecules/mol)  

         =  0.00925 mol

<h3>∴ the 5.57 x 10²¹ molecules of carbon dioxide ≡ 0.00925 mol</h3>
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A student obtains the following data: Mass of empty, dry graduated cylinder: 21.577 g Volume added of NaCl solution: 4.602 mL Ma
klasskru [66]

<u>Answer:</u> The density of NaCl solution is 3.930 g/mL

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 21.577 g

Mass of NaCl and cylinder combined, M = 39.664 g

Mass of NaCl, m_2 = (M-m_1)=(39.664-21.577)g=18.087g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of NaCl = 18.087 g

Volume of NaCl solution = 4.602 mL

Putting values in above equation, we get:

\text{Density of NaCl}=\frac{18.087g}{4.602mL}\\\\\text{Density of NaCl}=3.930g/mL

Hence, the density of NaCl solution is 3.930 g/mL

7 0
2 years ago
The smallest unit of an element that maintains properties of that element is called a(n)
Mazyrski [523]

Answer:

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Explanation:

7 0
3 years ago
15. What volume of CCI, (d = 1.6 g/cc) contain
anastassius [24]

Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

4 0
2 years ago
What is the formula for germicide that is sufficient to kill blood-borne pathogens?
Colt1911 [192]
It is 1 ounce. Bloodborne Pathogens can be transmitted when blood or body liquid from a tainted individual enters someone else's body by means of needle-sticks, human chomps, cuts, scraped areas, or through mucous films. Likewise, semen, vaginal discharges and salivation in dental methods are considered conceivably tainted body liquids.
7 0
3 years ago
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0
3 years ago
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