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kondor19780726 [428]
2 years ago
14

How many grams are in 3.45 moles CO2

Chemistry
1 answer:
Mamont248 [21]2 years ago
6 0
<h3>Hello There!!</h3>

<u> To Find:-</u>

Grams Present in 3.45 Moles of Carbon.

<u>We </u><u>Know:</u><u>-</u>

Molecular Mass of Carbon Dioxide= 44.01 grams/mol or 44 grams/mol

\text{Moles} = \frac{\text{Mass}}{\text{Molecular mass}}

\implies 3.45 \text{moles} =  \frac{ \text{Mass}}{44 \text{g/mol}}  \\  \\  \implies \text{Mass} = {44 \text{g/mol}} \times 3.45 \text{moles} \\  \\ \implies \text{Mass} = {44 \text{g/}  \cancel\text{ mol}} \times 3.45  \cancel\text{moles} \\  \\ \implies \text{Mass} = 44 \text{g} \times 3.45 \\  \\  \implies  \text{Mass} =  \red{151.8 \text{g}}

\therefore \text{Mass of 3.45 Moles of Carbon Dioxide is =  \red{151.8g}}

<h3>Hope This Helps!!</h3>
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Archy [21]

Answer:

the average kinetic energy of a system

Explanation:

the more kinetic energy in something, the hotter it is, the less kinetic energy in something, the cooler it is. gases have the most kinetic energy. solids have the least, but everything has kinetic energy

7 0
3 years ago
1375 J of heat energy are added to a 50.0 g sample of copper, which increases temperature from 22.8 C to 94.2 C. Calculate the s
sergeinik [125]

Answer:

0.385 J/g°C

Explanation:

Using the following formula, the specific heat of an elemental substance can be calculated:

Q = m × c × ∆T

Where;

Q = quantity of heat (J)

m = mass (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

Based on the given information, Q = 1375J, m = 50g, initial temperature = 22.8°C, final temperature = 94.2°C, c =?

Q = m × c × ∆T

1375 = 50 × c × (94.2 - 22.8°C)

1375 = 50 × c × 71.4

1375 = 3570c

c = 1375/3570

c = 0.3851

c = 0.385 J/g°C

3 0
3 years ago
What mass of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution?
inna [77]

Answer:

459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution

Explanation:

Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by the following expression:

Molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}

In this case:

  • Molarity: 1.56 M= 1.56 \frac{moles}{liter}
  • Number of moles of calcium chlorine= ?
  • Volume= 2.657 liters

Replacing:

1.56 M=\frac{Number of moles of calcium chlorine}{2.657 liters}

Solving:

Number of moles of calcium chlorine= 1.56 M* 2.657 liters

Number of moles of calcium chlorine= 4.14 moles

In other side, you know:

  • Ca: 40 g/mole
  • Cl: 35.45 g/mole

Then the molar mass of the calcium chloride CaCl₂ is:

CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole

Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?

mass=\frac{4.14 moles*110.9g}{1 mole}

mass= 459.126 g

<u><em>459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution</em></u>

3 0
3 years ago
A food substance kept at 0°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6
ZanzabumX [31]

Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

Given that:

time t_1 = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time t_2 = 10.6 hours

Temperature T_1 = 0° C = (0+273 )K = 273 K

Temperature T_2 = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since (\frac{k_2}{k_1}=\frac{t_2}{t_1})

Then we can rewrite the above expression as:

In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})

2.934 = \frac{E_a}{8.314}(\frac{30}{82719})

2.934 = \frac{30E_a}{687725.766}

30E_a = 2.934 *687725.766

E_a = \frac{2.934 *687725.766}{30}

E_a =67255.58 \ J/mol

E_a =67.2 \ kJ/mol

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3 years ago
How does the proximity to water affect temperature and precipitation in the tropical, temperate, and polar zones?
Zolol [24]

Answer:

Water heats and cools more slowly than landmasses.

Explanation:

because the coastal regions will stay cooler in summer and warmer in winter but creating a more moderate climate with a narrower temperature range.

5 0
3 years ago
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