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yuradex [85]
3 years ago
8

To complete the following chemical equation, what coefficients must be added? _____ Ca(OH)2 + _____ H3PO3 yields _____ Ca3 (PO3)

2 + _____ H2O
Chemistry
1 answer:
Katena32 [7]3 years ago
7 0
This is a neutralization reaction where calcium hydroxide and phosphoric acid are reactants whereas calcium phosphate and water are the products. any chemical reaction must be balanced for it to be considered correct. to balance this reaction we should put coefficients as follows 3Ca(OH)2+2H3PO3→Ca3(PO3)2+6H2O. this reaction is balanced since we 3
You might be interested in
All elements have a AHf, of ___
stellarik [79]

All elements have a AHf, of <u>0</u>

kJ/mole at standard temperature and pressure.

<h3>What is an element?</h3>

An element can be defined as a substance which cannot be split into two or more simpler forms by an ordinary chemical process

Below are the list of the first twenty elements:

  • Hydrogen
  • Helium
  • Lithium
  • Berylium
  • Boron
  • Carbon
  • Nitrogen
  • Oxygen
  • Fluorine
  • Neon
  • Sodium
  • Magnesium
  • Aluminum
  • Silicon
  • Phosphorus
  • Sulphur
  • Chlorine
  • Argon
  • Potassium
  • Calcium

So therefore, all elements have a AHf, of <u>0</u>

kJ/mole at standard temperature and pressure.

Learn more about elements:

brainly.com/question/11829854

#SPJ1

7 0
2 years ago
In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu
Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

7 0
2 years ago
a square boat made from iron has overall dimensions of 2.00 cm x 11.0 cm x 11.0 cm. it has a mass of 213 g. what is its density
Rus_ich [418]
Volume:

2.00 x 11.0 x 11.0 => 242 cm³

mass : 213 g

D = m / V

D = 213 / 242

D = 0.880 g/cm³

Answer B

hope this helps!
5 0
3 years ago
Read 2 more answers
II. Ionic Equations
mario62 [17]

Answer:

Complete ionic: \begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Net ionic: \begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

Explanation:

Start by identifying species that exist as ions. In general, such species include:

  • Soluble salts.
  • Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2. These three salts will exist as ions:

  • Each \rm AgNO_3\, (aq) formula unit will exist as one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion.
  • Each \rm CaCl_2 formula unit will exist as one \rm Ca^{2+} ion and two \rm Cl^{-} ions (note the subscript in the formula \rm CaCl_2\!.)
  • Each \rm Ca(NO_3)_2 formula unit will exist as one \rm Ca^{2+} and two \rm {NO_3}^{-} ions.

On the other hand, \rm AgCl is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2 (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each \rm AgNO_3\, (aq) formula unit will exist as only one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion. However, because the coefficient of \rm AgNO_3\, (aq)\! in the original equation is two, \!\rm AgNO_3\, (aq) alone should correspond to two \rm Ag^{+}\! ions and two \rm {NO_3}^{-}\! ions.

Do not rewrite the salt \rm AgCl because it is insoluble.

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of \rm Ca^{2+} and two units of \rm {NO_3}^{-}. Doing so will give:

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}.

Simplify the coefficients:

\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

7 0
2 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
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