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3 years ago

Which of the following compounds would contain an electrostatic attraction as bonds between ions?

Chemistry

1 answer:

4vir4ik [10]3 years ago

6 0

Answer:

iron sulfate

Explanation:

Electrostatic attraction as bonds between ions is characteristic of the electrovalent bond or the ionic bond

In this type of bond, there is electron transfer from one atom to another. The atom that looses electrons become positively charged while the atom that gains electrons becomes negatively charged.

In iron sulfate, there is electrostatic attraction between Fe II ions and sulphate ions, making iron sulfate an ionic compound.

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a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp

Drupady [299]

Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL$

Putting values in above equation, we get:

$1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148$

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_3PO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL$

Putting values in above equation, we get:

$3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M$

The concentration of the phosphoric acid solution is 1.172 M

3 0

3 years ago

Which statement describes both homogeneous mixtures and heterogeneous mixtures?

Mamont248 [21]

I believe the answer is D

4 0

3 years ago

After decaying for 48 hours, one-sixteenth (1/16) of the original mass of a radioisotope sample remains unchanged. What is the h

Hunter-Best [27]

The half-life of this radioisotope : 12 hr

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

t = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

t=48 hr

$\tt \dfrac{Nt}{No}=\dfrac{1}{16}$

The half-life :

$\tt \dfrac{1}{16}=\dfrac{1}{2}^{(48/t\frac{1}{2} )}\\\\(\dfrac{1}{2})^4=(\dfrac{1}{2})^{48/t\frac{1}{2}}\\\\4=48/t\frac{1}{2}\\\\t\frac{1}{2}=12~hr$

7 0

2 years ago

Which statement correctly describes how the parts of a flowering plant contribute to its survival? A. Roots prepare the plant fo

bekas [8.4K]

Answer:

what

Explanation:

Which statement correctly describes how the parts of a flowering plant contribute to its survival? A. Roots prepare the plant for seed production, while the stem releases pollen for fertilization. B. Sepals prepare the plant for pollination, while the leaves collect oxygen for photosynthesis. C. Petals are involved in the reproductive cycle, while stamens protect the plant from predators. D. Ovaries are involved in fertilization and seed production, while anthers and

8 0

2 years ago

Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the

fiasKO [112]

Answer:

a) Br2 + 3F2 → 2BrF3

b) Br2 is the limiting reactant.

c) There will be formed 86.3 grams of BrF3

d) There will remain 0.326 moles of F2 = 12.97 grams

Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

Br2 is the limiting reactant. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = 0.326 moles of F2 = 12.97 grams

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = 86.3 grams

7 0

2 years ago