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Talja [164]
3 years ago
6

This type of chemical bond forms when electrons are either gained or lost by an atom.

Chemistry
1 answer:
Ket [755]3 years ago
5 0
It's ionic because electrictrons are only lost or gained in ionic bonds
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How many grams are contained in a 0.183 mol sample of ammonium phosphate?
soldi70 [24.7K]

Molar mass of ( NH₄)₃PO₄ = 14.01×3 + 1.01×12 + 30.97 + 16.00×4 = 149.12 g/mol. Mass of 0.183 mol ...

7 0
2 years ago
HELP ME THIS IS 1 MONTH OVER DUE PLS HURRY
AVprozaik [17]
1false
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4 0
2 years ago
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Sulfuric acid (H₂SO₄), is formed when sulfur dioxide (SO₂) reacts with oxygen and water. Using the balanced chemical reaction ab
Lynna [10]

Answer:

8.5 mol H₂SO₄

Explanation:

It seems the balanced reaction the problem is referring to is absent, however the description matches the following balanced reaction:

  • 2SO₂ + O₂ + 2H₂O → 2H₂SO₄

Now we <u>can convert 8.5 moles of SO₂ into moles of H₂SO₄</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:

  • 8.5 mol SO₂ * \frac{2molH_2SO_4}{2molSO_2} = 8.5 mol H₂SO₄
5 0
3 years ago
What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?
deff fn [24]

Answer:

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

V₂ = 5.6 L;   T₂ = ?  

=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

3 0
3 years ago
Suppose you are titrating a sulfuric acid solution of unknown concentration with a sodium hydroxide solution according to the eq
Inga [223]

Answer:

M_{acid}=0.0444M

Explanation:

Hello,

In this case, for the given reaction:

H _2 S O _4 + 2 N a O H \rightarrow 2 _H 2 O + N a _2 S O_ 4

We find a 1:2 molar ratio between the acid and the base respectively, for that reason, at the equivalence point we find:

2*n_{acid}=n_{base}

That in terms of concentrations and volumes we can compute the concentration of the acid solution:

2*M_{acid}V_{acid}=M_{base}V_{base}\\\\M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}}=\frac{0.697M*28.07mL}{2*220.1mL}\\ \\M_{acid}=0.0444M

Best regards.

5 0
3 years ago
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