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2 years ago

PLEASE HELP

Chemistry

1 answer:

nata0808 [166]2 years ago

7 0

Answer:

Is this math? Cause as a fourth grader, I can do Algebra, but not this.

Explanation:

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What is a spectorscope?

vitfil [10]

Answer:

Explanation:

An apparatus for producing and recording spectra for examination.

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6 0

3 years ago

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Sakura tested the pH of several solutions. Which substances are acidic? Check all that apply. Cola-flavored soda: 2. 6 cranberry

Archy [21]

Cola-flavored soda: 2. 6 -Acidic

cranberry juice: 2. 9 -Acidic

dish soap: 7. 3 -Basic

pickle juice: 5. 5 -Acidic

seawater: 8. 0 -Basic

lye soap: 13. 0. -Basic

pH value tells us about the acidity and basicity of any solution. pH means potential hydrogen.

The pH scale is counted from 0 to 14 in which 7 is neutral and the value of pure water.

Below 7 are all acids and the above 7 are all basic.

Thus, all acidic substances are Cola-flavored soda: 2. 6, pickle juice: 5. 5, cranberry juice: 2. 9.

Learn more about pH, here:

brainly.com/question/491373

4 0

3 years ago

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mo

aleksandrvk [35]

Explanation:

Relation between temperature and activation energy according to Arrhenius equation is as follows.

k = $A exp^{\frac{-E_{a}}{RT}}$

where, k = rate constant

A = pre-exponential factor

$E_{a}$ = activation energy

R = gas constant

T = temperature in kelvin

Also,

$ln (\frac{k_{2}}{k_{1}}) = (\frac{-E_{a}}{R}) \times (\frac{1}{T_{2}} - \frac{1}{T_{2}})$

$T_{1}$ = $244^{o}C$ = (244 + 273) K = 517.15 K

$T_{2}$ = $324^{o}C$ = (597.15 + 273) K = 597.15 K

$k_{1}$ = 6.7 $M^{-1} s^{-1}$ , $k_{2}$ = ?

R = 8.314 J/mol K

$E_{a}$ = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows.

$ln (\frac{k_{2}}{6.7}) = (\frac{-71000}{8.314} \times (\frac{1}{597.15} - \frac{1}{517.15})$

= 2.2123

$\frac{k_{2}}{6.7} = exp(2.2123)$

= 9.1364

$k_{2} = 61 M^{-1}s^{-1}$

Thus, we can conclude that rate constant of this reaction is $61 M^{-1}s^{-1}$ .

3 0

4 years ago

Bismuth oxide reacts with carbon to form bismuth metal: Bi2O3(s) + 3C(s) → 2Bi(s) + 3CO(g) When 775 g of Bi2O3 reacts with exces

goldfiish [28.3K]

Answer:

(a) 3.33 mol Bi

(b) 140. g CO

Explanation:

Step 1: Convert 775g to moles

Bi Molar Mass - 208.98 g/mol × 2 = 417.98 g/mol

O Molar Mass - 16.00 g/mol × 3 = 48.00 g/mol

775g Bi₂O₃ ÷ 465.98 g/mol = 1.66316 mol Bi₂O₃

Step 2: Find the conversion from Bi₂O₃ to Bi

1 mole of Bi₂O₃ equals 2 moles of Bi

Step 3: Use Dimensional Analysis

1.66316 mol · $\frac{2 \hspace{2} mol \hspace{2} Bi}{1 \hspace{2} mol \hspace{2} Bi_2O_3}$ = 3.32632 mol Bi

3.32632 mol Bi ≈ 3.33 mol Bi (3 significant figures)

Step 4: Find the conversion from Bi₂O₃ to CO

1 mole of Bi₂O₃ equals 3 moles of CO

Step 5: Use Dimensional Analysis

1.66316 mol · $\frac{3 \hspace{2} mol \hspace{2} CO}{1 \hspace{2} mol \hspace{2} Bi_2O_3}$ = 4.98948 mol CO

Step 6: Find molar mass of CO and convert moles to grams

C - 12.01 g/mol

O - 16.00 g/mol

4.98948 mol CO · 28.01 g/mol = 139.775 g CO

139.775 g CO ≈ 140. g CO (3 significant figures)

6 0

4 years ago

The molar heat of fusion for water is 6.01 kJ/mol. How much energy must be added to a 75.0-g block of ice at 0°C to change it to

andreyandreev [35.5K]

Answer is: 25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.

7 0

3 years ago

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