Which of the following is the greatest speed: 20 m/s, 200 cm/s, or 0.2 km/ s

A mass of 5.4 grams of aluminum (al) reacts with an excess of copper (ii) chloride (cucl2) in solution, as shown below. 3cucl2 2

horrorfan [7]

Answer:

Mass of copper produced is 19.07g

Explanation:

Let's bring out the chemical equation;

3CuCl2 + 2Al → 2AlCl3 + 3Cu

3 : 2 : 2 : 3

Upon confirming that the reaction is indeed balanced, we can proceed.

The questions asks to calculate mass of Cu formed when a mass of 5.4g of Al is being used.

From the equation, what is the relationship between Al and Cu?

2 mol of Al would react to form 3 mol of Cu

Expressing this in terms of mass, we have;

mass = no. of moles * molar mass

Mass of Aluminium = 2 * 26.98 = 53.98 g

Mass of Copper = 3 * 63.546 = 190.638g

This means;

53.98 g of Al would react to form 190.638g of Cu

So how much Cu would form from 5.4 g of Al?

This leads us to;

53.98 = 190.638

5.4 = x

Upon cross multiplication, we are left with;

x = (190.638 * 5.4) / 53.98

x = 19.07 g

Mass of copper produced is 19.07g

8 0

2 years ago

An amount of solid barium chloride, 20.8 g, is dissolved in 100 g water in a coffee-cup calorimeter by the reaction: BaCl2 (s) 

mamaluj [8]

Answer : The enthalpy change during the reaction is -6.48 kJ/mole

Explanation :

First we have to calculate the heat gained by the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

m = mass of water = 100 g

c = specific heat = $4.04J/g^oC$

$T_{final}$ = final temperature = $26.6^oC$

$T_{initial}$ = initial temperature = $25.0^oC$

Now put all the given values in the above formula, we get:

$q=100g\times 4.04J/g^oC\times (26.6-25.0)^oC$

$q=646.4J$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles barium chloride = $\frac{\text{Mass of barium chloride}}{\text{Molar mass of barium chloride}}=\frac{20.8g}{208.23g/mol}=0.0998mole$

$\Delta H=-\frac{646.4J}{0.0998mole}=-6476.95J/mole=-6.48kJ/mole$

Therefore, the enthalpy change during the reaction is -6.48 kJ/mole

8 0

2 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

2 years ago

On a separate sheet of paper, write the formulas for the missing components of these neutralization reactions.

Fofino [41]

The formula of the missing component are a follows

HBr + KOH ------> KBr + H2O

H2SO4 + 2NH4OH--------> (NH4)2SO4 + 2H2O

2HNO3 + Mg(OH)2 ---------> Mg(NO3)2 + 2 H2O

4 0

2 years ago

Read 2 more answers

A species has the following MO configuration: (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)2. This substance is:_______.

Goshia [24]

Answer : The correct option is, (a) paramagnetic with two unpaired electrons.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 14 electrons present in the given configuration.

The molecular orbital configuration of molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.

Hence, the correct option is, (a) paramagnetic with two unpaired electrons.

3 0

2 years ago