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3 years ago

Factorf(x)=x² + 4x +3

Chemistry

1 answer:

My name is Ann [436]3 years ago

7 0

Answer:

f(x)=(x+1)(x+3)

Explanation:

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Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl

anzhelika [568]

Answer: ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

$\Delta H=-277.4kJ$

Therefore, the enthalpy change for this reaction is, -277.4 kJ

4 0

3 years ago

Which of the following are second-order reactions? a. C2H6 → 2 CH3 b. 2 N2O5 → 4 NO2 + O2 c. 2 HI → H2 + I2 d. 2 N2O → 2 N2 + O2

Klio2033 [76]

Answer:

The answer is b, c, d, e

Explanation:

b. 2 N2O5 → 4 NO2 + O2

r = k [N2O5]^2 --> Second-order regarding global reaction

c. 2 HI → H2 + I2

r = k [HI]^2 --> Second-order regarding global reaction

d. 2 N2O → 2 N2 + O2

r = k [N2O]^2 --> Second-order regarding global reaction

e. 2 NO2 → 2 NO + O2

r = k [NO2]^2 --> Second-order regarding global reaction

8 0

3 years ago

Kitty [74]

Answer:

0.14607220033346532

Explanation:

3 0

3 years ago

How many moles are in 24.75 g of H2O? What is given the conversation and the unknown?

riadik2000 [5.3K]

Answer:

1.373 mol H₂O

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

24.75 g H₂O

Step 2: Identify Conversions

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Convert

$24.75 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O} )$ = 1.37347 mol H₂O

Step 4: Check

We are given 4 sig figs. Follow sig fig rules and round.

1.37347 mol H₂O ≈ 1.373 mol H₂O

7 0

3 years ago

The chemical formula for ferric sulfate is Fe(SO4)3. Determine the following:

Lyrx [107]

Answer :

(a) The number of sulfur atoms are, $31.61\times 10^{23}$ .

(b) The mass of the mass of $Fe_2(SO_4)_3$ is, 1059.682 grams.

(c) The number of moles of $Fe_2(SO_4)_3$ is, $8.63\times 10^{-3}mole$

(d) The mass of the mass of $Fe_2(SO_4)_3$ is, $19.95\times 10^{-22}g$

Explanation :

(a) As we are given the number of moles of $Fe_2(SO_4)_3$ is, 1.75 mole. Now we have to calculate the number of sulfur atoms.

In the $Fe_2(SO_4)_3$ , there are 2 iron atoms, 3 sulfur atoms, 12 oxygen atoms.

As, 1 mole of $Fe_2(SO_4)_3$ contains $3\times 6.022\times 10^{23}$ number of sulfur atoms.

So, 1.75 mole of $Fe_2(SO_4)_3$ contains $1.75\times 3\times 6.022\times 10^{23}=31.61\times 10^{23}$ number of sulfur atoms.

The number of sulfur atoms are, $31.61\times 10^{23}$

(b) As we are given the number of moles of $Fe_2(SO_4)_3$ is, 2.65 mole. Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$\text{Mass of }Fe_2(SO_4)_3=2.65mole\times 399.88g/mole=1059.682g$

The mass of the mass of $Fe_2(SO_4)_3$ is, 1059.682 grams.

(c) As we are given the mass of $Fe_2(SO_4)_3$ is, 3.45 grams. Now we have to calculate the moles of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$3.45g=\text{Moles of }Fe_2(SO_4)_3\times 399.88g/mole$

$\text{Moles of }Fe_2(SO_4)_3=8.63\times 10^{-3}mole$

The number of moles of $Fe_2(SO_4)_3$ is, $8.63\times 10^{-3}mole$

(d) As we are given the formula unit of $Fe_2(SO_4)_3$ is, 3. Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

As we know that 1 mole of $Fe_2(SO_4)_3$ contains $6.022\times 10^{23}$ formula unit.

Formula used :

$\text{Formula unit of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}$

$3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}$

$\text{Moles of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole$

Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$\text{Mass of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole\times 399.88g/mole=19.95\times 10^{-22}g$

The mass of the mass of $Fe_2(SO_4)_3$ is, $19.95\times 10^{-22}g$

6 0

3 years ago

Factorf(x)=x² + 4x +3​

Factorf(x)=x² + 4x +3