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BaLLatris [955]
3 years ago
13

Which type of bond is salt

Chemistry
2 answers:
tigry1 [53]3 years ago
8 0

Answer:

An Ionic Bond

Explanation:

Salt compounds are formed by the "donating" of electrons.

Alex17521 [72]3 years ago
5 0

Answer:

NaCl, or Salt, is an ionic bond.

Explanation:

Ionic bonds are formed from one metal and one nonmetal, Sodium being the metal and Chlorine being the nonmetal.

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What is the silver ion concentration in a solution prepared by mixing 425 mL 0.397 M silver nitrate with 427 mL 0.459 M sodium p
Lisa [10]

Answer:

0 M is the silver ion concentration in a solution prepared mixing both the solutions.

Explanation:

molarity=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}

Moles of silver nitrate = n

Volume of the solution = 425 mL = 0.425  L (1 mL = 0.001 L)

Molarity of the silver nitrate solution = 0.397 M

n=0.397 M\times 0.425 L=0.1687 mol

Moles of sodium phosphate = n'

Volume of the sodium phosphate solution = 427 mL = 0.427  L (1 mL = 0.001 L)

Molarity of the sodium phosphate solution = 0.459 M

n'=0.459 M\times 0.427 L=0.1960 mol

3AgNO_3+Na_3PO_4\rightarrow Ag_3PO_4+3NaNO_3

According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :

\frac{1}{3}\times 0.1687 mol=0.05623 mol of sodium phosphate

This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.

So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.

0 M is the silver ion concentration in a solution prepared mixing both the solutions.

4 0
3 years ago
Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm
pav-90 [236]

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

7 0
3 years ago
In the simple act of lighting a match, chemical energy stored in the match head is transformed into heat and light. What is true
harina [27]

The head of a matchstick has a great deal of chemical energy stored in it, including combustible substances that produce a flame when rubbed against a suitable surface. ... As the combustible materials burn, some of the chemical energy is transformed into heat energy, and some is transformed into light energy. Hope this helps

5 0
3 years ago
Read 2 more answers
3. (10 Points) You buy a helium balloon of 14 Liters for your friend's birthday and
vesna_86 [32]

Answer: picture shows work for # 1,2,4,5,7

Explanation:

number 3: as the pressure in the volume decreases, the volume increases causing it to expand and eventually blow.

number 6: because the temperature and the amount of gas don’t change, these terms don’t appear in the equation. What Boyle’s law means is that the volume of a mass of gas is inversely proportional to its pressure. This linear relationship between pressure and volume means doubling the volume of a given mass of gas decreases its pressure by half.

hope this helps :))

7 0
3 years ago
In the equation CH4 + 2O2 --> 2H2O + CO2 What is the mass of CO2 produced when 35g of O2 reacts?
Anestetic [448]

Answer:

24.06 g of CO₂

Explanation:

The balanced equation for the reaction is given below:

CH₄ + 2O₂ —> 2H₂O + CO₂

Next, we shall determine the mass of O₂ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Finally, we shall determine the mass of CO₂ produced by the reaction of 35 g of O₂. This can be obtained as follow:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Therefore, 35 g of O₂ will react to produce = (35 × 44)/64 = 24.06 g of CO₂.

Thus, 24.06 g of CO₂ were produced from the reaction.

8 0
2 years ago
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