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1 year ago

Calculate the mass of 2.50 mol of CH,OH(1). Show your work. Use the appropriate

Chemistry

1 answer:

Sophie [7]1 year ago

7 0

Answer:

80.1 grams

Explanation:

Find the molar mass of CH3OH first by using the periodic table values.

12.011 g/mol C + (1.008*3 g/mol H) + 15.999g/mol O + 1.008 g/mol H

=32.042 so that is the molar mass

Now that you have 2.50 moles of CH3OH, you can calculate the mass in g

2.50molCH3OH * (32.042g CH3OH / 1 mol CH3OH) = 80.105

32.042g / 1 mol is the same as 32.042 g/mol

Since there are 3 sig figs in the problem (2.50 has 3 sig figs), you round to 80.1 g CH3OH

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0.01 M HCl solution has a pH of 2. Suppose that during the experiment, both the universal pH indicator and the cabbage indicator

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It matches the universal pH indicator and is indicating the proper pH

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2 years ago

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How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?

zaharov [31]

Answer:

$m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

$2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2$

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

$m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

$4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2$

Best regards!

7 0

1 year ago

Consider the following oxides: SO2, Y2O3, MgO, Cl2O, and N2O5. How many are expected to form acidic solutions in water? Consider

Snowcat [4.5K]

Answer:

Three of the five oxides are expected to form acidic solutions in water

Explanation:

We have different types of oxides : Acidic oxides, Basic oxides, Amphoteric oxides, Peroxides and Higher oxides.

Only acidic oxides will dissolve in water to give an acidic solution.

Considering the given oxides carefully,

SO2 will dissolve in water to produce H2SO3 which is acidic.

Y2O3 will dissolve in water to produce Yttrium(III) hydroxide which is basic.

MgO will dissolve in water only to produce Mg(OH)2 which is also basic.

Cl2O dichlorine mono oxide will dissolve in water to produce HClO which is acidic.

N2O5 will dissolve in water to produce HNO3 which is also acidic.

5 0

1 year ago

Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57

vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0

1 year ago

A 251 g strip of glass wool is used to insulate a reaction flask. During the reaction the temperature of the glass wool increase

Ivahew [28]

Answer:

8.9 KJ

Explanation:

Given data:

Mass of strip = 251 g

Initial temperature = 22.8 °C

Final temperature = 75.9 °C

Specific heat capacity of granite = 0.67 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 75.9 °C - 22.8 °C

ΔT = 53.1 °C

Q = 251 g × 0.67 j/g.°C × 53.1 °C

Q = 8929.8 J

Jolue to KJ.

8929.8J ×1 KJ / 1000 J

8.9 KJ

8 0

1 year ago