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3 years ago

Which has a greater volume, 10 grams of water or 10 grams of acetone?

Chemistry

1 answer:

WINSTONCH [101]3 years ago

6 0

Density is defined as the ratio of mass to the volume.

Density = $\frac{Mass}{Volume}$ (1)

Mass of water = 10 grams

Mass of acetone = 10 grams

Density of water = 1 $\frac{g}{cm^{3}}$

Density of acetone = 0.7857 $\frac{g}{cm^{3}}$

Put the value of density of water and its mass in equation (1)

1 $\frac{g}{cm^{3}}$ = $\frac{10 g}{volume}$

Volume of water = 10 $cm^{3}$

Put the value of density of acetone and its mass in equation (1)

0.7857 $\frac{g}{cm^{3}}$ = $\frac{10 g}{volume}$

Volume of acetone = 12.72 $cm^{3}$

Thus, volume of acetone is more than volume of water because the density of acetone is lower.

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You have two buffered solutions. Buffered solution 1 consists of 5.0 M HOAc and 5.0 M NaOAc; buffered solution 2 is made of 0.05

andre [41]

Answer:

Explanation:

The Henderson-Hasselbalch equation relates pH to the concentrations of an weak acid-base conjugate pair as follows:

pH = pKa + log([A⁻]/[HA])

For solution 1, the pH may be expressed as follows:

pH = pKa + log(5.0M/5.0M) = pKa

For solution 2, pH may be expressed as follows:

pH = pKa + log(0.050M/0.050M) = pKa

Thus, the pH values are equal to the pKa in both cases and are the same.

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3 years ago

Fill in the missing blank 2C4H6 + _______ → 8CO2 + 6H2O

Len [333]

the missing blank is filled by 11 O2
that is
= 2C4H6 + 11O2 = 8 CO2 + 6H2O

C4H6 reacted with oxygen (O2 )through the process of combustion to form carbon iv oxide (CO2) and water (H2O) . 11 infront of O2 is to make sure the molecules of O2 is balanced in both reactant side and the product side

3 0

3 years ago

How many grams are in 0.75 moles of SO2

lana66690 [7]

<h3>Answer:</h3>

48 g SO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.75 mol SO₂

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of S - 32.07 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of SO₂ - 32.07 + 2(16.00) = 64.07 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.75 \ mol \ SO_2(\frac{64.07 \ g \ SO_2}{1 \ mol \ SO_2})$
Multiply/Divide: $\displaystyle 48.0525 \ g \ SO_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

48.0525 g SO₂ ≈ 48 g SO₂

7 0

3 years ago

How many protons, neutrons, and electrons are there in an atom of (a) 10B5, (b) 36Ar, (C) 85Sr38, and (d) carbon-Il?

astraxan [27]

Answer:

Here's what I get.

Explanation:

The isotopic symbol for an atom is

$\rm _{Z}^{M}A$

The atomic number (Z) is the number of protons (p) in the nucleus.

In a neutral atom, the number of electrons (e) equals the number of protons. n = e

The mass number (A) is the number of protons plus the number of neutrons (n).

A = p + n or

n = A - p

(a) $\rm _{5}^{10}B$

p = 5

e = 5

n = 10 - 5 = 5

(b) $\rm _{18}^{36}Ar$

p = 18

e = 18

n = 36 - 18 = 18

(c) $\rm _{38}^{85}Sr$

p = 38

e = 38

n = 85 - 38 = 47

(d) $\rm _{6}^{11}C$

p = 5

e = 6

n = 11 - 6 = 5

5 0

4 years ago

The complete combustion of copper(I) sulphide is according to the following equation:

MaRussiya [10]

Answer:

a) 7.94 x 10²³ molecules, b) 5.62 g SO2, c) 1.97 L

Explanation:

a) We need to first convert Cu2S to moles. Since molar mass of Cu2S is 159.14 g/mol, 14.0 g = 0.0880 mol. Using molar ratios (3 mol O2/2 mol Cu2S, 0.0880 mol of Cu2S = 0.132 mol O2. Since 1 mol contains 6.02 x 10²³ molecules, 0.132 mol O2 = 7.94 x 10²² molecules O2.

b) Since the molar ratios of Cu2S to SO2 is 1:1, 0.0880 mol of Cu2S produces 0.0880 mol SO2. To convert mol to grams, we use the molar mass of SO2 (64.06 g/mol) to figure out that 0.0880 mol SO2 = 5.63 g SO2.

c) At STP, 1 mol occupies 22.4 liters. 0.0880 mol SO2 x 22.4 L/1 mol = 1.97 L

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5 0

3 years ago