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3 years ago

How many moles of Ca atoms are in 1 mol of CaSO4?

Chemistry

1 answer:

Galina-37 [17]3 years ago

7 0

Answer:

6.02 x 10²³ atoms

Explanation:

Given parameters:

Number of moles CaSO₄ = 1 mole

Unknown:

Number of Ca atoms in the given compound = ?

Solution:

The given compound is:

CaSO₄

1 mole of CaSO₄ is made up of 1 mole of Ca atoms

Now;

1 mole of any substance contains 6.02 x 10²³ atoms

1 mole of Ca atoms will also contain 6.02 x 10²³ atoms

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Which orbital is partially filled in the Zirconium (ZI) atom?

Nikolay [14]

Answer:

4d orbital.

Explanation:

Hello!

In this case, since zirconium's atomic number is 40, we fill in the electron configuration up to 40 as shown below:

$1s^2,2s^2,2p^6, 3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^2$

Thus, the orbital 4d is partially filled.

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2 years ago

How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The o

SVETLANKA909090 [29]

Answer:

Explanation:

8 0

2 years ago

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Purification of nickel can be achieved by electrorefining nickel from an impure nickel anode onto a pure nickel cathode in an el

Alexxandr [17]

Answer: 530 hours

Explanation:

The reduction of Nickel ions to nickel is shown as:

$Ni^{2+}+2e^-\rightarrow Ni$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole of Nickel

1 mole of Nickel weighs = 58.7 g

Given quantity = 18.0 kg = 18000 g (1kg=1000g)

58.7 g of Nickel is deposited by 193000 C of electricity

18000 g of Nickel is deposited by = $\frac{193000}{58.7}\times 18000=59182282.8C$ of electricity

$Q=I\times t$

where Q= quantity of electricity in coloumbs = 59182282.8C

I = current in amperes = 31.0 A

t= time in seconds = ?

$59182282.8C=31.0A\times t$

$t=1909105.9sec$

(1h=3600 sec)

$t=530hours$

Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A

3 0

2 years ago

What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

REY [17]

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

5 0

2 years ago

Which element would have the lowest electronegativity? (1 point)

irinina [24]

The element that will have the lowest electronegativity is an element with a small number of valence electrons and a large atomic radius.

Electronegativity of an element is the ability or power of that element in a molecule to attract electrons to its Valence electrons. The following are the properties of electronegativity:

It increases across a period from left to right of the periodic table,

It decreases down the periodic table groups

Group 1 elements are the least (lowest) electronegative elements. These elements have the lowest valence electrons with a large atomic radius.

Group 7 elements are the most electronegative elements.

Atomic radius of elements increase down a group because of a progressive increase in the number of shells occupied by electrons which increases the size. But it decreases across a period because electrons are accommodated within the same shell leading to greater attraction by the protons in the nucleus.

Learn more about electronegativity of elements here:

brainly.com/question/20348681

6 0

2 years ago