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3 years ago

Why is gold usually found in its pure form

Chemistry

1 answer:

attashe74 [19]3 years ago

5 0

gold is usually found in pure form because it is not reacting with other chemicals naturally.

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How many significant figures are in 3246.90890

Nonamiya [84]

It think it has exactly 9

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3 years ago

Picture a neutral S atom. This neutral atom will have

asambeis [7]

Answer:

has 6 valence electrons as its in group 6

the atom will gain electrons

the charge will be 2- as it will gain 2 electrons to be a full configuration

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3 years ago

Can you define what chemestry is? (Maybe in a easy way)

Mamont248 [21]

Answer:

Explanation:

Chemistry, the science that deals with the properties, composition, and structure of substances (defined as elements and compounds), the transformations they undergo, and the energy that is released or absorbed during these processes.

7 0

3 years ago

Read 2 more answers

A certain compound has the percent composition (by mass) 85.63% C and 14.37% H. The molar mass of the compound is 42.0 g/mol. Ca

AlekseyPX

Answer:

The molecular formula is C3H6

Explanation:

Step 1: Data given

Suppose the compound has a mass of 100 grams

The compound contains:

85.63 % C = 85.63 grams C

14.37 % H = 14.37 grams H

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles

Moles = grams / molar mass

Moles C = 85.63 grams / 12.01 g/mol

Moles C = 7.130 moles

Moles H = 14.37 grams / 1.01 g/mol

Moles H = 14.2 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 7.130 moles / 7.130 moles = 1

H = 14.2 moles / 7.130 moles = 2

The empirical formula is CH2

The molar mass of CH2 = 14 g/mol

Step 4: Calculate molecular formula

We have to multiply the empirical formula by n

n = 42 / 14 = 3

n*(CH2) = C3H6

The molecular formula is C3H6

8 0

3 years ago

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K

charle [14.2K]

Answer: The equilibrium constant for the total reaction is $4.09\times 10^{-6}$

Explanation:

We are given:

$K_{c_1}=0.282\\\\K_{c_2}=41$

We are given two intermediate equations:

Equation 1: $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282$

The expression of $K_{c_1}$ for the above equation is:

$K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}$ .......(1)

Equation 2: $H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41$

The expression of $K_{c_2}$ for the above equation is:

$K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}$

$41=\frac{[HI]^2}{[H_2][I_2]}$ ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

$(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}$

Now, dividing expression 1 by expression 2, we get:

$\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}$

$\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}$

The above expression is the expression for equilibrium constant of the total equation, which is:

$2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c$

Hence, the equilibrium constant for the total reaction is $4.09\times 10^{-6}$

8 0

3 years ago