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3 years ago

I cant seem to figure out ANY of these... help when you can pls :.)

Chemistry

1 answer:

Nonamiya [84]3 years ago

7 0

The first one is some reaction with water even I am studying the same

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FeS2 + 5 O2 2Fe2 + 4SO2 How many moles of FeO are formed from 0.57 moles of O2?

bagirrra123 [75]

Answer:

0.228 mol FeO

Explanation:

2FeS2 + 5 O2----> 2FeO + 4SO2

from reaction 5 mol 2 mol

given 0.57 mol x

x = 0.57*2/5 = 0.228 mol FeO

3 0

3 years ago

chlorine gas forms went to green out of the shower and electron what type of bonding is present in chlorine gas

sattari [20]

The answer is Chemical bounding

4 0

3 years ago

Which of these items is made up of one or more molecules?<br> Gold<br> Hydrogen<br> Salt<br> Neon

Nesterboy [21]

2 one hope it’s right

3 0

3 years ago

How many atoms of Mg are present in 97.22 grams of Mg?

Solnce55 [7]

Answer:

2.439 x 10E24 atoms

Explanation:

from

N=n x NA where

N- number of atoms

n-number of moles and NA is the avogadro's number

6 0

3 years ago

If the ka of a monoprotic weak acid is 5.4 × 10-6, what is the ph of a 0.14 m solution of this acid?

Aleks [24]

pH = 2.1

Let $HA (aq)$ resembles the acid; as a weak acid (a small value of $K_{a}$ ) $HA$ would partially dissociate to produce protons $H^{+}$ and $A^{-}$ , its conjugate base. Let the final proton concentration (i.e., $[H^{+}]$ ) be $x$ . (Apparently $x \ge 0$ ) Construct the following RICE table:

$\left\begin{array}{cccccc}\text{R}&HA(aq)&\rightleftharpoons&H^{+}(aq) &+ &A^{-}(aq)\\\text{I} & 0.14 \; \text{M} & &\\\text{C}&-x \; \text{M}& &+x \; \text{M} & & +x \; \text{M}\\E & (0.14 - x)\; \text{M} & & x \; \text{M} & &\+x \; \text{M}\end{array}\right$

By definition, (all concentrations are under equilibrium condition)

$\left\begin{array}{ccc}K_{a}&=&[H^{+}] \cdot [A^{-}] / [HA]\\&=&x^{2} /(0.14 - x)\end{array}\right$

It is given that

$K_a = 5.4 \cdot 10^{-6}$

Equating and simplifying the two expressions gives a quadratic equation; solve the equation for $x$ gives:

$x^2 = 5.4 \cdot 10^{-6} \cdot (14 - x) \\x^2 + 5.4 \cdot 10^{-6} \cdot 14 \cdot x - 5.4 \cdot 10^{-6} \cdot 14 = 0 \\x = 0.0087 \; \text{M} \; (x \ge 0)$

The pH of a solutions equals the opposite of the logarithm of its proton concentration to base 10; thus for this particular solution

$\text{pH} = -\text{ln(}[H^{+}]\text{)} / \text{ln(}10\text{)} = 2.1$

7 0

4 years ago