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3 years ago

1) A light bulb takes in 30 of energy per second. It transfers 3j as use

Chemistry

1 answer:

natta225 [31]3 years ago

8 0

Answer:

$\boxed{\text{10 \%}}$

Explanation:

The formula for efficiency is

$\begin{array}{rcl}\text{Efficiency} & = & \dfrac{\text{useful energy out}}{\text{energy in}} \times 100 \,\% \\\\\eta & = & \dfrac{w_{\text{out}}}{w_{\text{in}} } \times 100 \,\%\\\end{array}$

Data:

Useful energy = 3 J

Energy input = 30 J

Calculation:

$\begin{array}{rcl}\eta & = & \dfrac{\text{3 J}}{\text{30 J}} \times 100 \,\%\\\\\eta & = & 10 \, \%\\\end{array}\\\text{ The efficiency is }\boxed{\textbf{10 \%}}$

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7. Balance the following reaction under basic conditions: Al (s) + Cr2O72- (aq) -> Al3+ (aq) + Cr3+ (aq)

MArishka [77]

The free energy change of the reaction; Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s) is calculated to be -443.83KJ/mol.

For the reaction shown in question 7, we can divide it into half equations as follows;

Oxidation half equation;

6 Al (s) -------> 6Al^3+(aq) + 18e

Reduction half equation;

3Cr2O7^2-(aq) + 42H^+(aq) + 18e -----> 6Cr^3+(aq) + 21H2O(l)

The balanced reaction equation is;

6Al(s) + 3Cr2O7^2-(aq) + 42H^+(aq) -----> 6Al^3+(aq) + 6Cr^3+(aq) + 21H2O(l)

The E° of this reaction is obtained from;

E° anode = -0.04 V

E°cathode = +1.50 V

E° cell = +1.50 V - (-0.04 V) = 1.54 V

Given that;

ΔG° = -nFE°cell

n = 3, F = 96500, E°cell = 1.54 V

ΔG° = -(3 × 96500 × 1.54)

ΔG° = -443.83KJ/mol

Learn more: brainly.com/question/967776

4 0

3 years ago

If you have significant concerns about a child's development, you should A. sensitively present your concerns to the family and

Veronika [31]

I think its answer choice A because it would better to talk to their parents first and not take matters into your own hands.

6 0

3 years ago

Read 2 more answers

What masses of monobasic and dibasic sodium phosphate will you use to make 250 mL of 0.1 M sodium phosphate buffer, pH = 7?

Oduvanchick [21]

Answer:

Mass of monobasic sodium phosphate = 1.857 g
Mass of dibasic sodium phosphate = 1.352 g

Explanation:

The equilibrium that takes place is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)

To solve this problem we use the Henderson–Hasselbalch (H-H) equation:

pH = pka + $log\frac{[A^{-} ]}{[HA]}$

In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21

If we use put data in the H-H equation, and solve for [HPO₄⁻²], we're left with:

$7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}]$

From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M

We replace the value of [HPO₄⁻²] in this equation:

0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M

1.616 * [H₂PO₄⁻] = 0.1 M

[H₂PO₄⁻] = 0.0619 M

With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 0.0619 M = 0.1 M

[HPO₄⁻²] = 0.0381 M

With the concentrations, the volume and the molecular weights, we can calculate the masses:

Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.

mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g

mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g

5 0

4 years ago

Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n-, where n is typically greater than 10,00

pentagon [3]

Answer:

7.68 × 10²⁴

Explanation:

Step 1: Calculate the mass of 1 molecule of the monomer CH₂CHCN

We will get the mass of the monomer by adding the masses of the elements.

mCH₂CHCN = 3 × mC + 3 × mH + 1 × mN

mCH₂CHCN = 3 × 12.01 amu + 3 × 1.01 amu + 1 × 14.01 amu = 53.07 amu

Step 2: Convert the mass of the monomer to grams

We will use the conversion factor 1 amu = 1.66 × 10⁻²⁴ g

53.07 amu × 1.66 × 10⁻²⁴ g/1 amu = 8.81 × 10⁻²³ g

Step 3: Calculate "n"

We will divide the mass of the polymer by the mass of the monomer.

n = 676.8 g / 8.81 × 10⁻²³ g = 7.68 × 10²⁴

6 0

3 years ago

N2 + 3H2 —> 2NH3

aliina [53]

Explanation:

47.0g is the mass of ammonia

8 0

3 years ago