Write a balanced chemical equation for the formation of magnesium oxide, mg0, from magnesium and oxygen

Which label belongs in the area marked X? can reproduce by budding can reproduce by fragmentation with regeneration reproduce in

marusya05 [52]

Answer:

Nucleus.

Explanation:

The sponge is a multicellular organism that consists of pores that allows water to move through the body. Sponges belong to the kingdom Animalia and phylum Porifera. Sponges can reproduce by budding. Sponges are placed in kingdom Animalia because they are unable to make their own food, made of more than one cell, and absence of cell wall.

4 0

3 years ago

The heat of combustion of ethane is -337.0kcal at 25 degrees celsius. what is the heat of the reaction when 3g of ethane is burn

Ghella [55]

Answer:

Q = -33.6kcal .

Explanation:

Hello there!

In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:

$Q=n*\Delta _cH$

Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:

$n=3g*\frac{1mol}{30.08g}=0.1mol$

Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:

$Q=0.1mol*-337.0\frac{kcal}{mol}\\\\Q=-33.6kcal$

Best regards!

8 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

3 years ago

Predict how many grams of KCI is produced from 40 grams of K?

Nutka1998 [239]

Explanation:

firstly find for the molar mass of kcl and molar mass of k

and then

molar mass of k = x

molar mass of kcl= 40

cross mutiply and then simplify you will get your answer

5 0

3 years ago

Why Group 18 elements on the periodic table rarely form compounds

lawyer [7]

Most elements on group 18 are the Noble Gases. They already have a complete last level with 8 electrones. Actually they can form compounds but only on the lab and they will not even last half a second.

4 0

3 years ago