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3 years ago

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The differences between two molecules include the type of sugar that forms a section of the molecules and the identity of one of

the four nitrogenous bases that make up another section of the molecules. These two molecules are —

1 answer:

klio [65]3 years ago

3 0

The answer is nucleic acids

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What is the unit of temperature on a solubility graph

Andrej [43]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound

Softa [21]

Answer: Ca3(PO4)2

Explanation:

We assume that the sample is 100.0 g.

The sample contains 38.7 g of Ca, 19.9 g of P, and 41.2 g of O as the proportions show.

Then we can calculate the number of moles of each component (n = m/atomic mass).

The number of moles of Ca = 38.7/40.078 = 0.96 mol.

The number of moles of P = 19.9/30.97 = 0.64 mol.

The number of moles of O = 41.2/15.99 = 2.75 mol.

Now, we can get the molar ratios of different components (Ca, P, and O) in the sample by dividing the number of moles of each component by the lower number of moles, that we should divide the number of moles by (0.64).

Ca: P: O = (0.96/0.64) : (0.64/0.64) : (2.75/0.64) = 1.5 : 1 : 4.

To avoid the fraction of the ratios, we can multiply all ratios by 2.0.

Now, the ratio of Ca : P : O will be 3 : 2 : 8.

That main the empirical formula of the compound is Ca3P2O8 which can be expressed as calcium phosphate (Ca3(PO4)2).

8 0

2 years ago

Density is the amount of mass per unit volume. Calculate the density of an object that has a mass of 10 Kg and has a volume of 5

Gnoma [55]

Answer: 2 kg/L or 2 g/mL

Explanation:

litres are a cubic measure. “Cubic litres” is redundant at best, potentially confusing.

4 0

2 years ago

You placed 6.35 g of a mixture containing unknown amounts of BaO(s) and MgO(s) in a 3.50-L flask containing CO₂(g) at 30.0°C and

adelina 88 [10]

Mass of BaO in initial mixture = 3.50g

Explanation:

Let mass of BaO in mixture be x g

mass of MgO in mixture be (6.35 - x) g

Initially CO_2

Volume = 3.50 L

Temp = 303 K

Pressure = 750 torr = 750 / 760 atm

Applying ideal gas equation

PV = nRT

n = PV / RT

(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303

(n)_CO_2 = 0.139 mole

Finally; mole of CO_2

n= PV /RT

((245/760) *3.5) / 303* 0.0821

(n)_CO_2 = 0.045 mole

Mole of CO_2 reacted = 0.139 - 0.045

=0.044 mole

BaO + CO_2 BaCO_3

Mgo + CO_2 MgCO_3

moles of CO_2 reacted = ( moles of BaO + moles of MgO)

moles of BaO in mixture = x / 153 mole

moles of MgO in mixture = 6.35 - x mole / 40

Equating,

x/ 153 +6.35/40 = 0.094

= x/153 + 6.35 / 40 - x/40 =0.094

= x (1/40 - 1153) = (6.35/40 - 0.094)

= x * 10.018464

= 0.06475

mass of BaO in mixture = 3.50g

5 0

2 years ago

Ethanol is being used as an additive to

CaHeK987 [17]

Answer:

1.Begin energy in kilojoules.

2.Use a conversion factor in convert kilojoules to joules .

3.Use a conversion factor top convert

3 0

2 years ago