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Verdich [7]
3 years ago
7

5)a strip of paper has a length of 14cm what is the lengh of the strip of paper in mm, and in km

Chemistry
1 answer:
I am Lyosha [343]3 years ago
7 0
A) 1 cm ---------- 10 mm
    14 cm ---------- ?

14 x 10 / 1 => 140 mm
-------------------------------------------------------

b) 1 cm ----------- 0.00001 km
    14 cm ---------- ?

14 x 0.00001 / 1 => 0.00014 km

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How many moles are in 48.0g of H2O2?
Tju [1.3M]

Answer:

1.41 moles H2O2(with sig figs)

Explanation:

okay so what is the molar mass of H2O2= (1.008 g/mol)2+(16.00g/mol)2= (2.016+ 32.00) g/ mol

= 34. 02 g/mol

48.0g H2O2* 1 mol H2O2/ 34.02 g H2O2= 1.41 mol H2O2

3 0
3 years ago
Gold has a molar (atomic) mass of 197 g/mol. consider a 2.47 g sample of pure gold vapor. (a) calculate the number of moles of g
adell [148]
N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol

n = 2.47 / 197 
n = 0.01253 moles.
I'm sure you wanted to ask more than this. Just put some comments in. I can do the same.
3 0
3 years ago
Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati
worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4  +  Cl-

Firstly, we calculate the number of moles of  the ammonia  as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

                 NH4+      H2O     ⇄  NH3        H3O+

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b    = K w

 

, where  K w- the self-ionization constant of water, equal to  

10 ^-14  at room temperature

This means that you have

K a = K w.K b   = 10 ^− 14 /1.8 * 10^-5 =  5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈  0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0
3 years ago
What is the molar mass of KOH?
Julli [10]
<h3>Answer:</h3>

56.11 g/mol

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Compound] KOH

<u>Step 2: Identify</u>

[PT] Molar Mass of K - 39.10 g/mol

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

[PT] Molar Mass of H - 1.01 g/mol

<u>Step 3: Find</u>

39.10 + 16.00 + 1.01 = 56.11 g/mol

5 0
3 years ago
Cansjhdhsbdhajajajhsbsbshs hahahah she jansbsbsns haha be safe
V125BC [204]

Answer:

wait what?

Explanation:

brainliest pls

4 0
3 years ago
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