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gogolik [260]
3 years ago
7

Given the following unbalanced equation:

Chemistry
1 answer:
pshichka [43]3 years ago
4 0

Answer:

11.84 moles

Explanation:

We first balance the chemical equation

CoCl_{2} +  F_{2}  ------>   CoF_{2} + Cl_{2}

Both number of elements in the reactant side is equal to elements on the product side thus this equation is already balanced

Using mole ratio to find moles of COF_{2}

1 mole of CoCl2 reacts with 1 mole of F2 to produce 1 mole of CoF2 and ! mole of Cl2

Mole CoCl2 = Mole CoF2 = 11.84 moles

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The Molar mass(M.w) in g/mol of 6.3 grams of an ideal gas that placed in 5.0 L tank at ST
BigorU [14]

Answer:

28 g/mol, N2

Explanation:

Given data:

Volume of gas = 5.0 L

Mass of gas = 6.3 g

Pressure = 1 atm

Temperature = 273 K

Molar mass of gas = ?

Solution:

We will calculate the density first.

d = mass/ volume

d = 6.3 g/ 5.0 L

d = 1.26 g/L

Molar mass:

d = PM/RT

M = dRT/P

M =  1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm

M = 28 g/mol

Molar mass of N₂ is 28 g/mol thus given gas is N₂.

7 0
3 years ago
The hypothetical elements shown here (figures a–d) do not include hydrogen or helium. Which element would you expect to bond cov
Marizza181 [45]
The answer is "Figure a" i did it on plato lol
4 0
3 years ago
Write each mixed number as as a decimal. Use bar notation if the decimal is repeating.
aniked [119]

Number 1 is 4.875 Number 2 is 3.47 repeating

6 0
3 years ago
How many grams of Co3+ are present in 2.34 grams of cobalt(III) nitrite?
Assoli18 [71]

Answer:

m_{Co^{3+}}=0.563gCo^{3+}

Explanation:

Hello there!

In this case, since these mole-mass relationships are understood in terms of the moles of the atoms forming the considered compound, we first realize that the chemical formula of the cobalt (III) nitrate is Co(NO₃)₃ whereas there is a 1:1 mole ratio of the cobalt (III) ion (molar mass = 58.93 g/mol) to the entire compound. In such a way, we first compute the moles of the salt (molar mass = 58.93 g/mol) and then apply the aforementioned mole ratio to obtain the grams of the required cation:

m_{Co^{3+}}=2.34gCo(NO_3)_3*\frac{1molCo(NO_3)_3}{244.95 gCo(NO_3)_3} *\frac{1molCo^{3+}}{1molCo(NO_3)_3} *\frac{58.93gCo^{3+}}{1molCo^{3+}} \\\\m_{Co^{3+}}=0.563gCo^{3+}

Best regards!

4 0
3 years ago
100 POINTS! Final Honor Activity Question
castortr0y [4]

The change in temperature had the greatest effect at changing the volume of the balloon.

<h3>What are the gas laws?</h3>

The gas laws are used to describe the parameters that has to do with gases.

Given that;

P1 = 98.5 kPa

T1 = 18oC or 291 K

V1 =  74.0 dm3

P2 =  7.0 kPa

V2 = ?

T2 = 18oC or 291 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V2= P1V1T2/P2T1

V2 =  98.5 kPa *  74.0 dm3 * 291 K/ 7.0 kPa * 291 K

V2 = 1041.3 dm3

When;

V1 = 1041.3 dm3

T1 = 291 K

V2 = ?

T2 = 80oC or 353 K

V1/T1 = V2/T2

V1T2 = V2T1

V2 = V1T2/T1

V2 = 1041.3 dm3 * 353 K/291 K

V2 = 1263 dm3

The change in temperature had the greatest effect at changing the volume of the balloon.

Given that

V1 =  100 cm^3

T1 = 273 K

P1 = 1.01 * 10^5 Pa

V2 = ?

P2 =  3.00 x 10^-4 Pa

T2 = -180oC or 255 K

V2= P1V1T2/P2T1

V2 =  1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K

V2 = 3.14 * 10^10 cm^3

Learn more about gas laws:brainly.com/question/12669509

#SPJ1

7 0
2 years ago
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