Calculating Density Warm Up

The smallest unit that can exist as an element and still have the properties of that element is a(n):

Yuri [45]

The smallest unit that can exist as an element and still have proprieties of that element is an Atom

8 0

2 years ago

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Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?

lina2011 [118]

Answer: $k_a$ for $C_6H_5NH_3^+$ at 25°C is $0.588\times 10^{-4}$

Explanation: We are given $k_b$ of $C_6H_5NH_2$ at 25°C which is $1.7\times 10^{-9}$

To calculate the $k_a$ of $C_6H_5NH_3^+$ , we use the formula:

$k_w=k_a\times k_b$

$k_w\text{ at }25^o=1\times 10^{-14}$

Putting values in above equation, we get:

$1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}$

7 0

3 years ago

Perform the following mathematical operation and express the results to the correct number of significant figures. (0.102 x 0.08

Kobotan [32]

Answer:

2.26

Explanation:

First you should know what are significant figures,

1 . All non - zero digits are significant.

2 . The zero between two non- zero are significant.Ex 302 has 3 significant figures.

3. The zero before the decimal and before any non-zero digit is non significant. Example : 0.003 has only 1 significant figure.

4. The zero after non zero are non - significant . But the zeros after the decimal point are significant.

300 has only 1 significant figure

300.0 has 4 significant figure

The least number of significant figures present in the number should be there in the final answer of the calculation.

Look at the number having minimum significant digits :

It is 3 (every number has 3- significant figures ) So the answer should also contain 3 - significant figures.

$\frac{0.102\times 0.0821\times 273}{1.01}$

First solve the numerator part

$\frac{2.285}{1.01}$

$=2.2635$

Round off this number to 3 significant figures.

Answer is = 2.26

It has 3 - significant digit since all the digits are non-zero.

5 0

2 years ago

What is the percentage composition of CF4

adoni [48]

The percentage composition of CF4 is %C = 13.64% and %F = 86.36%

The solution are as follows:
Molar mass of CF4 = 88
molar mass of C = 12 
molar mass of F = 4x19 = 76 

% C = 12/88 x 100 = 13.64% 
% F = 76/88 x 100 = 86.36%

8 0

3 years ago

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The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

Answer: The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

Explanation:

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

4 0

2 years ago