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3 years ago

In comparing 1 mole of carbon atoms to one mole of magnesium atoms, which statement is TRUE?

Chemistry

1 answer:

Gemiola [76]3 years ago

3 0

Answer:

Explanation:

A) The molecular mass of a compound is the atomic mass of this compound but in grams, we know that the magnesium has an atomic mass of 24 u and the carbon has and atomic mass of 12 u, so the molecular mass of magnesium is 24g and the molecular mass of carbon is 12g, so a mole of magnesium has a greater mass.

B) A mole of any given substance is the avogadro number, 6.023 10^23 molecules, so a mole of magnesium and a mole of carbon has the same amount of atoms.

C) The molecular mass of a compound is the atomic mass of this compound but in grams, we know that the magnesium has an atomic mass of 24 u and the carbon has and atomic mass of 12 u, so the molecular mass of magnesium is 24g and the molecular mass of carbon is 12g, so a mole of magnesium has a greater mass.

D) The molecular mass of a compound is the atomic mass of this compound but in grams, we know that the magnesium has an atomic mass of 24 u and the carbon has and atomic mass of 12 u, so the molecular mass of magnesium is 24g and the molecular mass of carbon is 12g, so a mole of magnesium has a greater mass.

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A mixture of oxygen, hydrogen, and nitrogen exerts a total pressure of 378 kPa. If the partial pressures of oxygen and hydrogen

sertanlavr [38]

What's the relationship between total and partial pressure? The total pressure is the sum of the parcial pressures!

So for us, it would be:

378= 212+101+x

where x is the parcial pressure of nitrogen.

Now we count:
378= 212+101+x
378=313+x
378-313=x
65=x

So the parcial pressure exerted by nitrogen is 65!

8 0

3 years ago

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The second step is to calcul

Free_Kalibri [48]

Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.

2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O

1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O

Empirical formula
H2SO4

8 0

2 years ago

Read 2 more answers

How do you convert moles of a compound to mass?

laiz [17]

Answer:

Divide the mass of the compound in grams by the molar mass you just calculated. The answer is the number of moles of that mass of compound

Explanation:

3 0

3 years ago

Akosua has observed that anytime she drinks sobolo, her stool appears dark green. On one occasion

Anna007 [38]

Answer:

i. Sobolo is a Ghanian drink that is produced from red hibiscus flower that has an average pH of 6.7

It contains cyanidin and anthocyanins, which is a red pigment that is red in an acidic medium and changes green when introduced in a basic medium that has a high pH

The pH at the rectum of the digestive system = 5 to 8 (Slightly basic)

Therefore, what made the stool of Akosua green is that the sobolo drink changes to green in basic solution

ii. The stool which appeared green because she took sobolo turn into bright red upon mixing with the acidic WC water because of the presence of anthocyanins in sobolo, it turns red in an acidic medium

iii. Sobolo which turns green, or blue in a basic medium and red in an acidic medium can be used as a litmus solution to test the pH of a given substance

Explanation:

Sobolo or soobolo in Ghana is a name for the Hibiscus tea or tisane, which is made from calyces of the hibiscus plant, and has a sour (tangy) taste and appears bright red in color

8 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.