The weight of Patterson on the moon will be ..... This is 6th grade science by the way

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
delta U is change in internal energy 

delta U = q + w 
q is heat and w work done 
here work was done by the system means energy leaving the system so w is negative 

delta U = q + w 

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ 

q = m x c x delta T 

7211 J = 80.0 g x c x (225-25) °C 

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)

5 0

3 years ago

Estimate ΔH for the reaction using bond dissociation energies from Table 7.1. Give your answer in kcal. C6H12O6 has five C−C bon

Nat2105 [25]

The equation for the photosynthesis reaction in which carbon dioxide and water react to form glucose is . The hear reaction is the difference between the bond dissociation energies in the products and the bond dissociation energies of the reactants

The reactant molecules have 12 C = O, 12 H - O bonds while the product molecules have 5 C - C, 7 C – O, 5 H – O, and 6 O = O bonds. The average bond dissociation energies for the bonds involved in the reaction are 191 for C = O, 112 for H – O, 83 C –C, 99 C – H, 86 C – O, 119 O = O.

Substitute the average bond dissociation energies in the equation for and calculate as follows

= [12 (C=O) + 12 (H-O)] – [5(C-C) + 7(C-H) + 7 (C-O) + 5(H-O) + 6(O=O)]

= [12x191 kcal/mol + 12x112 kcal//mol] – [5x83 kcal/mol + 7x99 kcal/mol + 7x86 kcal/mol + 5x112 kcal/mol + 6x119 kcal/mol]

= 3636 kcal/mol – 2984 kcal/mol = 652 kcal/mol x 4.184 Kj/1kcal = 2.73x10^3 kJ/mol

So, enthalpy change for the reaction is 652 kcal/mol or 2.73x10^3 kJ/mol

5 0

3 years ago

Hydrogen and oxygen react chemically to form water how much water would form if 14.8grams of hydrogen reacted with 34.8 grams of

pishuonlain [190]

Answer:

There will be formed 39.1935 grams H2O formed

Explanation:

Step 1: The balanced equation

2H2 + 02 → 2H20

Step 2: Given data

mass of hydrogen = 14.8 grams

Molar mass of hydrogen = 2.02 g/mole

mass of oxygen = 34.8 grams

Molar mass of oxygen = 32 g/mole

Step 3: Calculate moles

moles = mass / Molar mass

moles of hydrogen = 14.8g/ 2.02 g/mole = 7.33 moles

moles of oxygen = 34.8g / 32g/mole = 1.0875 moles

For 2 moles hydrogen consumed, we need 1 mole of oxygen.

This means oxygen is the limiting reagens and will be consumed completely. Hydrogen is the reactant in excess, there will remain 5.155 moles of hydrogen

Step 4: Calculate moles of H2O

We see that for 2 moles of H2 consumed, there is needed 1 mole of O2, to produce 2 moles of H2O.

For 1.0875 moles of oxygen consumed, there will be produced 2.175 moles of H2O

Step 5: Calculate mass of water

Mass of H2O = moles of H2O * Molar mass of H2O

Mass of H2O = 2.175 moles * 18.02 g/moles 39.1935 grams

There will be formed 39.1935 grams H2O formed

4 0

3 years ago

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

2 years ago

which characteristics of an element can be determined by considering only the element’s specific location on the periodic table?

Nadusha1986 [10]

Answer:

Gas, Solid, or liquid

Explanation:

5 0

2 years ago

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