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2 years ago

- a) write the formula unit of calcium fluoride.

Chemistry

1 answer:

san4es73 [151]2 years ago

6 0

Answer:

Practice

Look at the pictures and write about Faris's day. Use time words

35words 40words * 50 words ***

TUD

Writing checkout

Olused capital letters at the beginning of sentences. Olsusea finger espace between e

the pictures (I will describe it by words)first picture the boy returns from school the sec picture is I return and eat with the family and the third picture is I played with my sister and the fourth is I watch TV with my siblings and the last picture is I take the book and study ,in order write a story using this pictures that I describe it.

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How can phisical change what a subtance?

solmaris [256]

Physical change can change a substance by affecting the form of a chemical change.
Examples of physical changes:
Something cut (Paper)
Molded (Bread)
Boiled (Water)
Mixed (Berries and strawberries)
HOPE THIS HELPS YOU! ^_^

3 0

3 years ago

How many moles of CaO will be produced from 23.9 g of Ca?<br>2Ca(s) + O2(g) – 20:0(3)<br>mol

inessss [21]

Answer:

0.6 moles of CaO will produced.

Explanation:

Given data:

Mass of calcium = 23.9 g

Moles of CaO produced = ?

Solution:

Chemical equation:

2Ca + O₂ → 2CaO

Number of moles of calcium:

Number of moles = mass/ molar mass

Number of moles = 23.9 g / 40 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of calcium and CaO.

Ca : CaO

2 : 2

0.6 : 0.6

0.6 moles of CaO will produced.

4 0

3 years ago

How does the sky appear blue? (BONUS QUESTION)!!!!

crimeas [40]

This was an answer i found on google from Nasa

6 0

3 years ago

Read 2 more answers

Tin has ten stable isotopes. The heaviest, 124Sn, makes up 5.80% of naturally occuring tin atoms. How many atoms of 124Sn are pr

matrenka [14]

The average atomic mass of Sn is 118.71 g/mol

the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = $6.023X10^{23}X0.691=4.16X10^{23}$

the percentage of heaviest Sn is 5.80%

So the total atoms of $Sn^{124}$ = 5.80% X $4.16X10^{23}$

Total atoms of $Sn^{124}$ = $2.41X10^{22}$ atoms

the mass of $Sn^{124}$ will be = $\frac{2.41X10^{22}X124}{6.023X10^{23}}=4.96g$

5 0

3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago