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3 years ago

X2o3 express you answer as a whole number

1 answer:

4 0

2.40 - 1.68 =0.72 g of oxigen
moles = 0.72/16 g/mol=0.045

moles x = 1.68/ 55.9=0.03

0.03/0.03 = 1 = x
0.045 / 0.03 = 1.5 = O

to get whole numbers multiply by 2

x2O3

X2O3 +3 CO = 2 X + 3 CO2

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For the oxidation-reduction reaction equation given here 4ga+p4 indicate how many electrons are transferred in the formation of

sergij07 [2.7K]

The reaction given is:

4Ga + P4 ---> 4GaP

The oxidation number of the reactants is zero, because they are pure elements.

The P in compounds may have oxidation states 3- or 5-. Gallium may only have oxidation state 3+.

Then, to be neutral in GaP the oxidation states are 3+ for Ga and 3- for P.

And the transference of electrons can be see in this oxidation - reduction equations:

Ga (0) - 3 e- ----> Ga (3+)

P (0) + 3e- ---> P (3-)

So, for one formula unit, 3 electrons have been transfered from each Ga atom to P atom to form one GaP unit.

Answer: 3 electrons.

3 0

3 years ago

In Experiment 9, a 1.05 g sample of a mixture of NaCl (MW 58.45) and NaNO2 (MW 69.01) is reacted with excess sulfamic acid. The

Kaylis [27]

Answer:

There is 76.6 mL of nitrogen collected

Explanation:

Step 1: Data given

Mass of the sample = 1. 05 grams

The sample is 40.00% by mass NaNO2

MW of NaCl = 58.45 g/mol

MW of NaNO2 = 69.01 g/mol

Temperature = 22.0 °C

Pressure = 750.0 mmHg = (750/760) atm

The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm

Step 2: Calculate mass of NaNO2

(40/100)*1.05 = 0.42 grams

Step 3: Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

Step 4: Calculate moles of N2

For 2 moles of NaNO2 we'll get 1 mol of N2

For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

Step 5: Calculate pressure of N2

P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa

Step 6: Calculate volume of N2

PV = nRT

⇒ P = the pressure of N2 = 0.96079

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = moles of N2 = 0.00304 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C = 295 Kelvin

V = (0.00304*0.08206*295)/0.96079

V = 0.0766 L = 76.6 mL

There is 76.6 mL of nitrogen collected

4 0

3 years ago

What is the density of an object with a mass of 1.663 g and a volume of 0.2009 mL

daser333 [38]

8.278 g/mL The definition of density is mass per volume. So what you need to do is divide the known mass by the known volume. So 1.663 g / 0.2009 mL = 8.27775 g/mL But you also have to keep track of significant figures. Since both 1.663 and 0.2009 have 4 significant figures each, you need to round the result to 4 significant figures. So 8.27775 g/mL = 8.278 g/mL

7 0

3 years ago

Read 2 more answers

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago

Porcentaje en masa formula y unidad

elena-14-01-66 [18.8K]

Explanation:

Expresa los gramos de soluto por cada 100 gramos de disolución. Porcentaje masa = masa de soluto___ x 100 masa de la disolución Cuando trabajamos con la masa, podemos sumar el soluto y el disolvente para obtener la disolución.

6 0

3 years ago