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3 years ago

Which statement best describes what happens when a person sees a blue ball?

Chemistry

2 answers:

Yuliya22 [10]3 years ago

8 0

The correct answer is B

Savatey [412]3 years ago

6 0

Actually, it was D. <span>blue light reflected from the ball enters human eyes. I just took the test.</span>

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How many significant figures are in 4.8cm?

kramer

Answer: 2. they are 4 and 8

Explanation: I took the exam. I hope I helped :)

4 0

3 years ago

Salts usually have a low melting point true or false

antoniya [11.8K]

Hello!

This is false!! It is the other way around. They usually have a very high melting point. For example table salt is around 800deg C which is approx just under 1500deg F.

Hope this helps. Thank you

3 0

3 years ago

A galvanic cell at a temperature of is powered by the following redox reaction: 2Cr3 + 3Ca Suppose the cell is prepared with in

stira [4]

Answer:

2.13 V

Explanation:

The balanced equation of the reaction his;

2Cr^3+(aq) + 3Ca(s) -----> 2Cr(s) + 3Ca^2+(aq)

Since this is a galvanic cell then E°cell must be positive. It implies that calcium will be the anode and chromium will be the cathode since calcium is ahead of chromium in the electrochemical series.

E°anode= -2.87 V

E°cathode= -0.74 V

E°cell= E°cathode -E°anode

E°cell= -0.74 -(-2.87)

E°cell = 2.13 V

8 0

3 years ago

How much heat energy is required to melt 75g of ice at 0°C?

Mnenie [13.5K]

The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:

Q = N x ΔH

where:

Q = heat

N = moles

ΔH = enthalpy

In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:

1. Calculate moles of water

2. multiply by the enthalpy of fusion

3. Convert kJ to J.

Step 1 : Calculate moles of water

$[ 75g ] x (\frac{1 mol}{18.02g} ) =$

Step 2 : Multiply by enthalpy of fusion

Q = N × ΔH = <em> [ Step 1 Answer ]</em> × 6.02 =

Step 3 : Convert kJ to J

$[ Step 2 Answer ] x (\frac{1000j}{1kJ} ) =$

Finally rounding to 2 sig figs (since 34°C has two sig figs) we get

Q Would Equal ____

4 0

3 years ago

A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .