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3 years ago

Why are some rocks more easily weathered that others

1 answer:

8 0

Because of their atom's composition. some are naturally packed closer together and stronger and more durable, while others tend to be more loose and breakable and/or weaker/less durable. also with how their atoms are composed is how the atoms interact with each other/ how they share or give or take electrons.

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Elements x and y form 2 binary compounds. In the first 14.0 g of x combines with 3.0 g of y. In the second, 7.00 g of x combine

Roman55 [17]

second compound

Let molar mass of x is = X

Let molar mass of y is = Y

Moles of x in second compound = Mass / molar mass = 7 / X

Moles of y in second compound = Mass / molar mass = 4.5 / Y

For second compound

7 / X : 4.5/ Y = 1:1

Therefore

X / Y = 7/4.5

Y / X = 4.5/ 7

The mass of x in first compound = 14g

moles of x in first compound = 14/X

Mass of y in first compound = 3

moles of y in first compound = 3 / Y

14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1

Thus molar ratio in first compound = moles of x / Moles of y = 3:2

Formula = x3y

6 0

3 years ago

How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----> K3PO4 + 3

8_murik_8 [283]

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

$\frac{3}{1} =\frac{y}{0.02256} \\$

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

5 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.