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3 years ago

How do you write out and balance calcium hydroxide reacts with hydrogen chloride to produce calcium chloride and water

Chemistry

1 answer:

Temka [501]3 years ago

5 0

Answer:

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Explanation:

Reactants;

Calcium hydroxide = Ca(OH)₂

Hydrogen chloride = HCl

Products;

Calcium chloride = CaCl₂

Water = H₂O

The reaction equation:

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

The reaction is a neutralization reaction where acids reacts with bases to produce salt and water only.

You might be interested in

Round off the following number to 3 digits 34,560

Vikki [24]

Answer:

3.46x10⁴

Explanation:

Hello,

In this case, we can see that the number 34,560 has five significant figures, it means that if we want to write it with three, we must take the 3, 4 and 5 only. Nevertheless, since the 6 after the five is greater than 5, we can round such five to 6, so we obtain:

346

However, the decimal places cannot get lost, therefore, we move the given thousand to the three, so the number turns out:

3.46x10⁴

Best regards.

6 0

3 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

3 years ago

What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor

ddd [48]

Answer:

$1.75272\ \text{m}^3$

Explanation:

The breakdown reaction of ozone is as follows

$CF_3Cl + UV \rightarrow CF_3 + Cl$

$Cl + O_3 \rightarrow ClO + O_2$

$O_3 + UV \rightarrow O_2 + O$

$ClO + O \rightarrow Cl + O_2$

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of $CF_3Cl$ = 15.5 g

M = Molar mass of $CF_3Cl$ = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = $62.363\ \text{L mmHg/K mol}$

Number of moles is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}$

$20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}$

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3$

For 20 cycles of the reaction the volume of the ozone is $1.75272\ \text{m}^3$ .

7 0

3 years ago

What is the daughter nuclide when 0-15 experiences positron emission?

Helen [10]

The daughter isotope (a decay product)of O-15 = N-15(Nitrogen 15)

<h3>Further explanation </h3>

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,

alpha α particles ₂He⁴
beta β ₋₁e⁰ particles
gamma particles γ
positron particles ₁e⁰

O-15 emits positron particles ₁e⁰, so the atomic number decreases by 1, the mass number is the same

Reaction

$\tt _8^{15}O\Rightarrow _7^{15}N+_1^0e$

The mass number of the daughter isotope = 15, atomic number = 7

If we look at the periodic system, the element with atomic number 7 is Nitrogen (N)

5 0

3 years ago

If the density of a liquid is 20g/mL and the volume is 70mL, what is the mass?

Archy [21]

Answer:

mass= volume ×density

20×70=1400 g

3 0

3 years ago