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3 years ago

A substance formed when two or more chemical elements are chemically bonded together?

Chemistry

1 answer:

Ronch [10]3 years ago

6 0

Answer: a piece of pure matter which we can see from our naked eye is termed as a substance

Explanation:

every substance has physical and chemical properties like;

diamond ,water ,pure sugar ,table salt e.t.c

an element is a substance made up of same type of atoms having same atomic number[protons],and cannot be easily decomposed in to a single substance by ordinary chemical reaction example,.....

gold,copper,zinc and some are in liquid form like mercury,bromine and some are in gases for example, hydrogen ,oxygen, nitrogen

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The reaction for photosynthesis producing glucose sugar and oxygen gas is:

Anvisha [2.4K]

1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

CO₂: 6 moles
H₂O: 6 moles
C₆H₁₂O₆: 1 mole
O₂: 6 moles

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

$moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}$

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

$moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}$

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

$mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}$

mass of glucose= 1.5 grams

Then, 1.5 grams of glucose is produced from 2.20 g of CO₂.

5 0

3 years ago

Which chemical out of the following 4 is the most dangerous? Glucose, benzene,acetone, ether

Lelechka [254]

Answer:

Explanation: hope this helps

Benzene

5 0

2 years ago

6. if a cross country runner travels at a speed of 10 mph and he runs for 2 hours, how many miles will he have run at the end of

storchak [24]

Explanation:

distance = speed x time

distance = 10 x 2

distance = 20miles

hope that helps :)

5 0

2 years ago

What is the reason behind that the melting and boiling point of iron is more than that of sodium?

sweet [91]

Answer:

As metals are giant lattice structures, the number of electrostatic forces to be broken is extremely large, and so metals have high melting and boiling points. This means that the melting point and boiling point of metals are more similar to those for ionic compounds than for covalent substances.

Explanation:

3 0

3 years ago

In the uncatalyzed reaction N2O4 (g) ⇌ 2 NO2 (g) the pressure of the gases at equilibrium are PN2O4 = 0.377 atm and PNO2 = 1.56

larisa86 [58]

Answer:

The pressures will remain at the same value.

Explanation:

A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.

While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.

Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).

The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.

Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.

8 0

3 years ago