What are the advantages of using resources such as geothermal energy, solar energy, and hydropower?

If the total mass of the products of a reaction is 90 gram, what was the total mass of the reactants

Marizza181 [45]

90 grams due to laws of conservation of mass. Output mass = input mass. Mass can never be created or destroyed.

4 0

3 years ago

A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug

OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 = log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] (1)

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M (2)

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = 0.035M

Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻

Moles of CH₃COOH before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles + 2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = 0.296

4 0

3 years ago

A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8

aivan3 [116]

Answer:

16.06 L was the initial volume of the balloon.

Explanation:

Initial moles of freon in ballon = $n_1=9.79 mol$

Initial volume of freon gas in ballon = $V_1=?$

Moles of freon gas added in the balloon = n = 3.50 mole

Final moles of freon in ballon = $n_2=n_1+n=9.79 mol+3.50 mol=13.29 mol$

Final volume of freon gas in ballon = $V_2=21.8 L$

Using Avogadro's law:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ ( at constant pressure and temperature)

$V_1=\frac{V_2\times n_1}{n_2}=\frac{21.8 L\times 9.79 mol}{13.29 mol}=16.06L$

16.06 L was the initial volume of the balloon.

4 0

3 years ago

I have no idea how to do this and need help

den301095 [7]

Answer:

60.55%

Explanation:

nO3=155/48

nO2 used: 3/2nO3=4.84375

percent yield: 4.84375/8=60.55%

5 0

3 years ago

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

6 0

3 years ago