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mixas84 [53]
2 years ago
11

What if there was an inverse relationship between the temperature and the volume? How do you think the world might be different?

How would this change our day-to-day lives?
Chemistry
1 answer:
kari74 [83]2 years ago
5 0

If there was an inverse relationship between the temperature and the volume, our daily lives change because in high temperature things will contract.

<h3>What if there was an inverse relationship between the temperature and the volume?</h3>

If there was an inverse relationship between the temperature and the volume then with increasing temperature decrease occur in the volume of a substance. If this type of relationship is present in the world, the objects will contract when the temperature is high and expand when the temperature is low which make the solid materials expand at winter and contract at summer season.

So we can conclude that if there was an inverse relationship between the temperature and the volume, our daily lives change because in high temperature things will contract.

Learn more about temperature here: brainly.com/question/25677592

#SPJ1

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How many atoms of carbon are found in a 30.0 gram sample of the element?
saveliy_v [14]

Answer:

1.50 × 10²⁴ atoms C

Explanation:

Step 1: Define

Molar mass of C - 12.01 g/mol

Avagadro's Number: 6.02 × 10²³ atoms, molecules, formula units, etc.

Step 2: Use Dimensional Analysis

30.0 \hspace{3} g \hspace{3} C(\frac{1 \hspace{3} mol \hspace{3} C}{12.01 \hspace{3} g \hspace{3} C} )(\frac{6.02(10)^{23} \hspace{3} atoms \hspace{3} C}{1 \hspace{3} mol \hspace{3} C} ) = 1.50375 × 10²⁴ atoms C

Step 3: Simplify

We are given 3 sig figs.

1.50375 × 10²⁴ atoms C ≈ 1.50 × 10²⁴ atoms C

3 0
3 years ago
A 25.0 mL aliquot of 0.0430 0.0430 M EDTA EDTA was added to a 56.0 56.0 mL solution containing an unknown concentration of V 3 +
Nady [450]

Answer:

Check the explanation

Explanation:

As we know the reaction of  EDTA and Ga^3+ and EDTA and V^3+

Let us say that the ratio is 1:1

Therefore, the number of moles of Ga^3+ = molarity * volume

                                    = 0.0400M * 0.011L

                                    = 0.00044 moles

Therefore excess EDTA moles = 0.00044 moles

Given , initial moles of EDTA  = 0.0430 M * 0.025 L

                                        = 0.001075

Therefore reacting moles of EDTA with V^3+ = 0.001075 - 0.00044 = 0.000675 moles

Let us say that the ratio between V^3+ and EDTA is 1:1

Therefore moles of V^3+ = 0.000675 moles

Molarity = moles / volume

                                    = 0.000675 moles / 0.057 L

                                    = 0.011 M (answer).

4 0
3 years ago
Ethyl alcohol, CH3CH2OH2, is misconception with octane, C8H18, but plenty alcohol, CH3CH2CH2CH2CH3OH3, is soluble in water only
emmasim [6.3K]
Because anymore water will breakdown the bonds of your Oh groups
3 0
3 years ago
N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?
just olya [345]

Answer:

\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     28.01               17.03

            N₂ + 3H₂ ⟶ 2NH₃

m/g:                          240.0

(a) Moles of NH₃

\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

(b) Moles of N₂

\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

(c) Mass of N₂

\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

7 0
3 years ago
Read 2 more answers
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
3 years ago
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