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3 years ago

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A. Metal ion can be pushed out of position B. Metal form covalent bond C. Metals are brittle

1 answer:

guapka [62]3 years ago

4 0

The answer is A. Good Luck!

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if 3.26 g of FeNO33 is dissolved in enough water to make exactly what is the molar concentration of nitrate ion g

Tanzania [10]

Answer:

0.404M

Explanation:

...<em>To make exactly 100.0mL of solution...</em>

Molar concentration is defined as the amount of moles of a solute (In this case, nitrate ion, NO₃⁻) in 1 L of solution.

To solve this question we need to convert the mass of Fe(NO₃)₃ to moles. As 1 mole of Fe(NO₃)₃ contains 3 moles of nitrate ion we can find moles of nitrate ion in 100.0mL of solution, and we can solve the amount of moles per liter:

<em>Moles Fe(NO₃)₃ -Molar mass: 241.86g/mol-:</em>

3.26g * (1mol / 241.86g) =

0.01348 moles Fe(NO₃)₃ * (3 moles of NO₃⁻ / 1mole Fe(NO₃)₃) =

<em>0.0404 moles of NO₃⁻</em>

In 100mL = 0.1L, the molar concentration is:

0.0404 moles of NO₃⁻ / 0.100L =

<h3>0.404M</h3>

5 0

2 years ago

How many moles are in 2.8x10^24 atoms of silicon?

sergejj [24]

Answer:

4.6 mol Si

General Formulas and Concepts:

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 2.8 × 10²⁴ atoms Si

[Solve] moles Si

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 2.8 \cdot 10^{24} \ atoms \ Si(\frac{1 \ mol \ Si}{6.022 \cdot 10^{23} \ atoms \ Si})$
[DA] Divide [Cancel like units]: $\displaystyle 4.64962 \ mol \ Si$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

4.64962 mol Si ≈ 4.6 mol Si

6 0

2 years ago

Phosphorus-32 has a half-life of 14.0 days. Starting with 4.00 g of 32P, how many grams will remain after 84.0 days ?

Lelu [443]

T₁=84 d
t₂=14 d
m₁=4 g

n=t₁/t₂
n=84/14=6

m₂=m₁/2ⁿ

m₂=4/(2⁶)=0.0625 g

0.0625 grams will remain after 84.0 days

3 0

3 years ago

Read 2 more answers

The molar solubility of zns is 1.6 Ã 10-12 m in pure water. Calculate the ksp for zns.

Oksi-84 [34.3K]

Answer:

The solubility product of zinc sulfide is $2.56\times 10^{-24}$ .

Explanation:

Solubility of the zinc sulfide = $S=1.6\times 10^{-12}$

$ZnS\rightleftharpoons Zn^{2+}+S^{2-}$

S S

The expression of solubility product is given by :

$K_{sp}=[Zn^{2+}]\times [S^{2-}]$

$K_{sp}=S\times S=S^2$

$=1.6\times 10^{-12}\times 1.6\times 10^{-12}=2.56\times 10^{-24}$

The solubility product of zinc sulfide is $2.56\times 10^{-24}$ .

5 0

3 years ago

In an experiment, a 0.4351 g sample of benzil (C14H10O2) is burned completely in a bomb calorimeter. The calorimeter is surround

andre [41]

Answer:

The enthalpy change during the reaction is -7020.09 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $994.1 J/^oC$

$T_{final}$ = final temperature = $27.60^oC$

$T_{initial}$ = initial temperature = $25.10^oC$

Now put all the given values in the above formula, we get:

$q=994.1 J/^oC\times (27.60-25.10)^oC$

$q= 2,485.25 J$

The heat gained by water present in calorimeter. = q'

$q'=mc\times (T_{final}-T_{initial})$

where,

q' = heat gained = ?

m = mass of water = $1.153\times 10^3 g$

c' = specific heat of water = $4.184 J/^oC$

$T_{final}$ = final temperature = $27.60^oC$

$T_{initial}$ = initial temperature = $25.10^oC$

$q'=1.153\times 10^3 g\times 4.184 J/^oC\times (27.60-25.10)^oC$

q ' = 12,060.38 J

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{Q}{n}$

where,

$\Delta H$ = enthalpy change = ?

Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J

Q = -14.54563 kJ

n = number of moles fructose = $\frac{\text{Mass of benzil}}{\text{Molar mass of benzil}}=\frac{0.4351 g}{210 g/mol}=0.002072 mole$

$\Delta H=-\frac{-14.54563 kJ}{0.002072 mole}=-7020.09 kJ/mole$

Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.

3 0

3 years ago