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3 years ago

GIVING BRAINLIEST FOR RIGHT ANSWERS PLS HELP

Chemistry

2 answers:

BigorU [14]3 years ago

5 0

Answer:

: Chemical change; Chemical property of heat of combustion

Explanation:

A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.

Example: Melting of ice

Physical property is defined as the property of a substance which becomes evident during physical changes. Example: Melting point

A chemical change is defined as a change in which a change in chemical composition takes place. A new substance is formed in these reactions.

Example: combustion

Chemical property is defined as the property of a substance which becomes evident during chemical changes. Example: Reactivity with other substances

Thus when a piece of coal and a piece of wood is burnt , there is formation of new substances and thus is a chemical change. Combustion is a chemical property as it is reactivity of a substance with oxygen.

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Anuta_ua [19.1K]3 years ago

4 0

Answer:

Answer: Chemical change; Chemical property of heat of combustion

Explanation:

A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.

Explanation:

You might be interested in

If you were to react 4.3 miles of magnesium phosphide how many moles of potassium would you need?

bixtya [17]

Answer:

25.8

Explanation:

Let's write the reaction between magnesium-phosphide and potassium:

Mg3P2 + K = Mg + K3P

And now let's balance this equation:

Mg3P2+6K=3Mg+2K3P

We see that the ratio of magnesium-phosphide and potassium is 1:6, which means that for every mole of magnesium-phosphide there need to be 6 moles of potassium.

Since we have 4.3 moles of Mg3P2, there need to be 6 • 4.3 = 25.8 moles of potassium.

5 0

3 years ago

What is the percent yield for the reaction below when

kumpel [21]

Answer:

Percent yield of reaction is<em> 150%.</em>

Explanation:

Given data:

Percent yield = ?

Actual yield of SO₃ = 586.0 g

Mass of SO₂ = 705.0 g

Mass of O₂ = 80.0 g

Solution:

Chemical equation:

2SO₂ + O₂ → 2SO₃

Number of moles of SO₂:

Number of moles = mass/ molar mass

Number of moles = 586.0 g/ 64.1 g/mol

Number of moles = 9.1 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 80.0 g/ 32g/mol

Number of moles = 2.5 mol

Now we will compare the mole of SO₃ with O₂ and SO₂.

SO₂ : SO₃

2 : 2

9.1 : 9.1

O₂ : SO₃

1 : 2

2.5 : 2×2.5 = 5

The number of moles of SO₃ produced by oxygen are less it will limiting reactant.

Theoretical yield of SO₃:

Mass = number of moles × molar mass

Mass = 5 mol × 80.1 g/mol

Mass = 400.5 g

Percent yield of reaction:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 586.0 g/ 400.5 g× 100

Percent yield = 1.5× 100

Percent yield = 150%

8 0

3 years ago

1. How many moles of carbon are in 2.25 * 10^22 atoms of carbon ?

tresset_1 [31]

Answer:

atoms of C? 2.25 x 1022 atoms C x 1 mole C = 0.037 mol C

Explanation:

How many moles of C are in 2.25 x 1022 atoms of C?

3 0

3 years ago

A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut

leonid [27]

Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

x(t)

200

We model this problem as

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

20 Lnit.

100 Lsol.

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

x(t) Lnit.

200 Lsol.

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dt = 1.2 − 0.03x,

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

x(0) Lnit.

200 Lsol.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

= e

0.03t

The solution is

x(t) = 1

µ(t)

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

0.03t

= Ce−0.03t +

1.2

0.03

−0.03t

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

3 0

3 years ago

What are the measurements for standard temperature and pressure (stp)? kp?

Alex787 [66]

Standard Temperature and Pressure (STP) are 273.15 K (Kelvin) and 101.3 KPa (kilopascals).

You can substitute 1 ATM (atmosphere) for the kilopascal measurement. It is discouraged to use 0°C for the temp equivalent since negative temperature numbers can't be used in gas equations..

8 0

4 years ago