Which of the following elements belong to the same pair?

New questionWhat mass of calcium chloride (CaCl₂) would beproduced from the reaction of 125.9 g of hydrochloriacid (HCI) with ex

marshall27 [118]

Answer:

191.6 g of CaCl₂.

Explanation:

What is given?

Mass of HCl = 125.9 g.

Molar mass of CaCl₂ = 110.8 g/mol.

Molar mass of HCl = 36.4 g/mol.

Step-by-step solution:

First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O.$

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

$125.9\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=3.459\text{ moles HCl.}$

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:

$3.459\text{ moles HCl}\cdot\frac{1\text{ mol CaCl}_2}{2\text{ moles HCl}}=1.729\text{ moles CaCl}_2.$

The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:

$1.729\text{ moles CaCl}_2\cdot\frac{110.8\text{ g CaCl}_2}{1\text{ mol CaCl}_2}=191.6\text{ g CaCl}_2.$

The answer would be that we're producing a mass of 191.6 g of CaCl₂.

4 0

1 year ago

For the reaction a2 + b2 → 2ab, ea(fwd) = 105 kj/mol and ea(rev) = 77 kj/mol. assuming the reaction occurs in one step, calculat

Lady bird [3.3K]

Answer is: 28 kJ.
Chemical reaction: A₂ + B₂ ⇄ 2AB.
Ea(forward) = 105 kJ/mol.
Ea(reverse) = 77 kJ/mol.
ΔH(reaction) = ?
The enthalpy change of reaction is the change in the energy of the reactants to the products.
ΔH(reaction) = Ea(forward) - Ea(reverse).
ΔH(reaction) = 105 kJ/mol - 77 kJ/mol.
ΔH(reaction) = 28 kJ/mol; this is endothermic reaction (ΔH > 0).

6 0

3 years ago

How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

zhenek [66]

Answer:

Grams of mercury= 0.06 g of Hg

Note: The question is incomplete. The complete question is as follows:

A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Explanation:

Since one fluorescent light bulb contains 4 mg of mercury,

15 such bulbs will contain 15 * 4 mg of mercury = 60 mg

1 mg = 0.001 g

Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.

Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.

5 0

3 years ago

PLEASE SOMEONE HELP ME WITH THIS!

Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877 and 4) 1.04x $10^{3\\}$

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x $\frac{1 L}{1000 mL}$ = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x $\frac{10 dg}{ 1 g}$ = 1.291 dg

For the next exercises, you need to follow some rules:

1. All numbers that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062 - 638.1848 = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.

1.04x $10^{3}$

4 0

3 years ago

A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of

inna [77]

The mass of the copper cylinder is 177065.856g

Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 $\frac{g}{c m^{3}}$

To find:

Mass of the copper cylinder

Step by Step by explanation:

Solution:

According to the formula, Mass can be calculated as

$\rho=\frac{m}{v}$ and from this

$m=\rho \times v$

Where, m=mass of the cylinder

$\rho$ =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

$v=\pi r^{2} h$

Where $\pi$ =3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, $v=3.14 \times 12^{2} \times 44$

v=3.14 \times 144 \times 44

$v=19895.04 \mathrm{cm}^{3}$

And we know that, $m=\rho \times v$

Substitute the known values in the above equation we get

$m=8.90 \times 19895.04$

m=177065.856g or 177.065kg

Result:

Thus the mass of the copper cylinder is 177065.856g

4 0

3 years ago