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3 years ago

Metric conversion 12 hg= 1.2 kg

Chemistry

2 answers:

Iteru [2.4K]3 years ago

8 0

1 Mg 1000 kg megagrams to kilograms table

1 dag 10 g dekagrams to grams table

1 kg 100 dag kilograms to dekagrams table

1 g 100 cg grams to centigrams table

1 hg 10000 cg hectograms to centigrams table

1 mg 0.1 cg milligrams to centigrams table

1 hg 100 g hectograms to grams table

1 hg 0.1 kg hectograms to kilograms table

1 ng 1.0E-9 g nanograms to grams table

1 dag 10000 mgExplanation:

kotegsom [21]3 years ago

5 0

Answer:

1 Mg 1000 kg megagrams to kilograms table

1 dag 10 g dekagrams to grams table

1 kg 100 dag kilograms to dekagrams table

1 g 100 cg grams to centigrams table

1 hg 10000 cg hectograms to centigrams table

1 mg 0.1 cg milligrams to centigrams table

1 hg 100 g hectograms to grams table

1 hg 0.1 kg hectograms to kilograms table

1 ng 1.0E-9 g nanograms to grams table

1 dag 10000 mg

Explanation:

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If Kc = 4.0×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 520 K , what is the value of Kp for this reaction at this temperature?

Flauer [41]

Here we have to get the $K_{p}$ of the reaction at 520 K temperature.

The $K_{p}$ of the reaction is 1.705 atm

We know the relation between $K_{p}$ and $K_{c}$ is $K_{p}=K_{c}(RT)^{N}$ , where $K_{p}$ = The equilibrium constant of the reaction in terms of partial pressure, $K_{c}$ = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.

Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅

Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole

The given value of $K_{c}$ is 4.0×10⁻²

The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.

On plugging the values in the equation we get,

$K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}$

Or, $K_{p}$ = 1.705 atm

Thus, the $K_{p}$ of the reaction is 1.705 atm

7 0

3 years ago

How many cm in 725 km?

tigry1 [53]

1 km is equal to 100,000 cm
$725km \times 100000cm \\ = 72500000 \: km .cm$
and you need to divide it to 1 km so that you can get rid of the km.
$= 72500000cm$

5 0

3 years ago

How much water would need to be added to 1092 mL of a 54.7 M NaCl solution to make a 0.25 M solution?

babunello [35]

Answer:

237.8L of water would need to be added.

Explanation:

The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).

Then figure out what information you have and what is being found. In this case:

M1 = 54.7 M

V1 = 1092 mL = 1.092 L

M2 = 0.25 M

V2 = unknown

Then solve the equation for whatever you are trying to find.

M1V1=M2V2

V2=M1V1/M2

Now you need to plug everything in.

V2=(54.7M*1.091L)/0.25M

V2=238.93L

That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.

238.93L - 1.091L = 237.8L

Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.

I hope this helps. Let me know if anything is unclear.

8 0

3 years ago

Twenty five grams of Iron 3 oxide react with an excess of carbon monoxide to form 15 g of Fe. Carbon dioxide is the other produc

densk [106]

<h3>Answer:</h3>

Theoretical mass = 17.42 g

Percent yield of Fe = 86.11%

<h3>Explanation:</h3>

The equation for the reaction between iron (iii) oxide and carbon monoxide is given by;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

We are required to calculate the theoretical yield and the percentage yield of Iron.

Step 1: Moles of iron (iii) oxide

Moles are given by dividing the mass of the compound by the molar mass.

Molar mass of Iron(iii) oxide = 159.69 g/mol

Moles of Iron(III) oxide = 25 g ÷ 159.69 g/mol

= 0.156 moles

Step 2: Moles of Iron produced

From the equation 1 mole of Iron(iii) oxide reacts to produce 2 moles of Fe.

Therefore, the mole ratio of Fe₂O₃ to Fe is 1 : 2.

Thus, moles of Fe = Moles of Fe₂O₃ × 2

= 0.156 moles × 2

= 0.312 moles

Step 3: Theoretical mass of iron produced

To calculate the mass of iron we multiply the number of moles of iron with the relative atomic mass.

Relative atomic mass = 55.845

Mass of iron = 0.312 moles × 55.845

= 17.42 g

Step 4: Percent yield of iron

% yield = (Actual mass ÷ Theoretical mass)×100

= (15 g ÷ 17.42 g) × 100 %

= 86.11%

7 0

3 years ago

What kind of car do Cherry's and Marcia's boyfriends ride around in

Bond [772]

I believe it was a blue mustang

6 0

3 years ago