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Nataly_w [17]
3 years ago
9

Why does it generally take more enthalpy to ignite a solid than a gas or liquid?

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

Answer:

It is due to the nature of the reactants

Explanation:

To ignite a solid, we require more heat component compared to liquids and gases. For ignition to occur, oxygen gas combines with a reactant in most cases.

Some factors affect the rate rate at which a chemical proceeds. One of the factors is the nature of reactants.

The solid phase is very slow while the gaseous phase is rapid and fast.

            solid phase < liquid phase <  gas phase

Gases are free and the molecules move in all direction. They easily combine and react very fast.

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Motivation that is physical and can be touched like a trophy or a medal is:
Talja [164]

Answer:

Tangible

Explanation:

Tangible alone means it is "perceptible by touch," meaning you physically observe the object with your sense of touch.

Having tangible motivation is like working for money or a trophy.

7 0
2 years ago
There are two steps in the extraction of copper metal from chalcocite, a copper ore. In the first step, copper(I) sulfide and ox
maxonik [38]

Answer:

2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO

Explanation:

Equation 1 should correctly be written as;

2Cu2S + 3O2-----> 2Cu2O + 2SO2

Equation 2 should be correctly written as;

2Cu2O + 2C -----> 4Cu + 2CO

The overall reaction equation is;

2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO

Note that species that are intermediates are cancelled out .

4 0
2 years ago
To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber
Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}

7 0
3 years ago
What does it mean for a strong base to be in equilibrium?
prisoha [69]

The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.

<h3>What is equilibrium?</h3>

Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.

A very high value of K indicates that at equilibrium most of the reactants are converted into products.

The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.

When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.

This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.

Hence, option A is correct.

Learn more about the equilibrium here:

brainly.com/question/23641529

#SPJ1

8 0
2 years ago
How many moles are present in 3.4 x 1023 atoms of Na?
Andrej [43]
<h3><u>moles of H2SO4</u></h3>

Avogadro's number (6.022 × 1023) is defined as the number of atoms, molecules, or "units of anything" that are in a mole of that thing. So to find the number of moles in 3.4 x 1023 molecules of H2SO4, divide by 6.022 × 1023 molecules/mole and you get 0.5646 moles but there are only 2 sig figs in the given so we need to round to 2 sig figs. There are 0.56 moles in 3.4 x 1023 molecules of H2SO4

Note the way this works is to make sure the units are going to give us moles. To check, we do division of the units just like we were dividing two fractions:

(molecules of H2SO4) = (molecules of H2SO4)/1 and so we have 3.4 x 1023/6.022 × 1023 [(molecules of H2SO4)/1]/[(molecules of H2SO4)/(moles of H2SO4)]. Now, invert the denominator and multiply:

<h3 />
7 0
2 years ago
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