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BARSIC [14]
2 years ago
10

Do any of the spheres exist on their own or are they intertwined?

Chemistry
1 answer:
Sauron [17]2 years ago
3 0

Answer:

show me a pic of the problem

Explanation:

show me a pic of the problem

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Ammonium phosphate nh43po4 is an important ingredient in many fertilizers. it can be made by reacting phosphoric acid h3po4 with
Tom [10]
The reaction between phosphoric acid and ammonia that produces ammonium phosphate can be written as follows:
3NH3 + H3PO4 ..................> (NH4)3PO4

From the periodic table:
molar mass of nitrogen = 14 grams
molar mass of hydrogen = 1 grams
molar mass of oxygen = 16 grams
molar mass of phosphorus = 30.9 grams

based on this:
molar mass of 3NH3 = 3 (14 + 3(1)) = 51 grams
molar mass of H3PO4 = 3(1) + 30.9 + 4(16) = 97.9 grams
molar mass of  (NH4)3PO4 = 3 (14 + 4(1)) + 30.9 + 4(16) = 54 + 30.9 + 64
                                            = 148.9 grams

Therefore, 97.9 grams of phosphoric acid is required to produced 148.9 grams of ammonium phosphate.
Thus, to know the mass of ammonium phosphate produced from 4.9 grams of phosphoric acid, we will simply use cross multiplication as follows:
amount of produced ammonium phosphate = (4.9 x 148.9) / 97.9 = 7.45 g
3 0
3 years ago
Read 2 more answers
Tartaric acid, C4H6O6, has the first ionization constant with the value: Ka1 = 9.20 × 10-4. Calculate the value of pKb for the c
nirvana33 [79]

Answer:

pKb = 10.96

Explanation:

Tartaric acid is a dyprotic acid. It reacts to water like this:

H₂Tart  +  H₂O  ⇄  H₃O⁺   +  HTart⁻         Ka1

HTart⁻  +  H₂O  ⇄  H₃O⁺   +  Tart⁻²           Ka2

When we anaylse the base, we have

Tart⁻²   +  H₂O  ⇄  OH⁻  +  HTart⁻       Kb1

HTart⁻  +  H₂O  ⇄  OH⁻  +    H₂Tart          Kb2

Remember that Ka1 . Kb2  = Kw, plus pKa1 + pKb2 = 14

Kb2 = Kw / Ka1  →   1×10⁻¹⁴ / 9.20×10⁻⁴  = 1.08×10⁻¹¹

so pKb = - log Kb2   → - log 1.08×10⁻¹¹ = 10.96  

7 0
3 years ago
Read 2 more answers
Help me with this question please.
Salsk061 [2.6K]

Explanation:

current =  \frac{voltage}{resistance}  \\ i = \frac{v}{ohm}  \\  i  = \frac{240}{38}  \\ i = 6.3158 \: amps

hope it helped

pls mark BRAINLIEST

thanks

8 0
3 years ago
A sample of CO2 weighing 86.34g contains how many molecules?
irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

86.34 g CO₂

<u>Step 2: Identify Conversion</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

<u />86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.18141 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
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