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2 years ago

20 What is the base-conjugate acid pair when a strong acid is added to methylamine, CH3NH2?

1 answer:

6 0

The base conjugate acid pair when strong acid is added to methylamine CH3NH2 is

base CH3NH2; conjugate acid: CH3NH3 +

Explanation

When a base CH3CH2 accept proton it form its conjugate acid which known as methyl ammonium ion. Therefore conjugate acid of CH3CH2 is CH3NH3+ ion

For example when CH3CH2 react with HCl

CH3CH2 +HCl → CH3CH3+ + Cl-

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If 32.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Arada [10]

Find moles of MgSO4.7H2O

molar mass = 246
so moles = 32 / 246 = 0.13 moles.

When heated, all 7 H2O from 1 molecule will be gone.
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18 = 16.38g

so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g

6 0

1 year ago

Metals with ________ electron configurations characteristically form diamagnetic, square planar complexes.

erastovalidia [21]

Answer:

Explanation:

Diamagnetism , paramagnetism and ferromagnetism are terms which are used to describe the magnetic activity of transition metals. These terms helps to know the response the metal will have when placed in a magnetic field. These activities can be discerned from the d-block electronic configuration. If the number of electrons are even, this means they all form a pair and the metal is diamagnetic. If otherwise, there is an unpaired electron which causes the paramagnetic activity of the metal in magnetic field.

To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes

8 0

2 years ago

Why is mining for coal a long-term concern?

Free_Kalibri [48]

A. Fossil fuels take a very long time to form so we mine it faster than it can replenish.

8 0

1 year ago

Read 2 more answers

A food product is being frozen in a system capable of removing 6000 kJ of thermal energy. The product has a specifi c heat of 4

lapo4ka [179]

Answer : The exit temperature of the product is, $-35.2^oC$

Explanation :

Total heat = Heat lost by liquid + Latent heat of fusion + Heat lost by frozen

$Q=m\times c_1\times (T_2-T_1)+m\times L_f+m\times c_2\times (T_4-T_3)$

where,

Q = Total heat = 6000 kJ

m = mass of product = 15 kg

$c_1$ = specific heat of liquid = $4kJ/kg^oC$

$L_f$ = latent heat of fusion = $275kJ/kg$

$c_2$ = specific heat of frozen = $2.5kJ/kg^oC$

$T_1$ = initial temperature of liquid = $2^oC$

$T_2$ = final temperature of liquid = $10^oC$

$T_3$ = initial temperature of frozen = ?

$T_4$ = final temperature of frozen = $2^oC$

Now put all the given value in the above expression, we get:

$6000kJ=[15kg\times 4kJ/kg^oC\times (10-2)^oC]+[15kg\times 275kJ/kg]+[15kg\times 2.5kJ/kg^oC\times (2-T_3)^oC]$

$T_3=-35.2^oC$

Thus, the exit temperature of the product is, $-35.2^oC$

7 0

1 year ago

What is the molar mass of 56.75 g of gas exerting a pressure of 2.87 atm on the walls of a 5.29 l container at 230 k?

laiz [17]

We first need to find the number of moles of gas in the container
PV = nRT
where;
P - pressure - 2.87 atm x 101 325 Pa/atm = 290 802.75 Pa
V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation
290 802.75 Pa x 5.29 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 230 K
n = 0.804 mol
the molar mass = mass present / number of moles
molar mass of gas = 56.75 g / 0.804 mol
therefore molar mass is 70.6 g/mol

8 0

1 year ago