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3 years ago

What is the enthalpy change when 49.4 ml of 0.430 m sulfuric acid reacts with 23.3 ml of 0.309 m potassium hydroxide?

1 answer:

7 0

Potassium hydroxide commonly known as caustic potash is an inorganic compound with a chemical formula KOH which is a strong base that is marketed in several forms.

H2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l) H°= –111.6 kJ/mol

The answer in this question is -0.402 kJ

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3(x - 2) = 5(x + 4)<br><br>

Tcecarenko [31]

Answer:

Uh first of all this is algebra but I'll answer this

First distribute the three and 5 (Multiply them by both terms inside parenthesis.

3x-6=5x+20

Then add like terms

8x=14

Divide 8 by 8 and 8 by 14

x = 14/8

Explanation:

5 0

3 years ago

Read 2 more answers

What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)

amid [387]

Answer:

$m=59.3gCO_2$

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

$M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol$

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

$mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol$

Finally, the mass by using the molar mass:

$m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2$

Best regards.

4 0

3 years ago

What is the mass in grams of 1.00 x 10 24 atoms of Mn?

Alex17521 [72]

<h3>Answer:</h3>

91.2 g Mn

<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
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

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

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 91.2321 \ g \ Mn$