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2 years ago

Can someone please answer this?

Chemistry

2 answers:

Blizzard [7]2 years ago

7 0

Answer:

Either A or B

Explanation: Because I learned it on Edge. :)

Lelechka [254]2 years ago

6 0

Answer:

B) go look on the periodic table for the atomic number

Explanation:

The atomic number is how many protons is in an atom

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A balloon is inflated such that a radius is 7 cm what is the volume of air inside the balloon in milliliters

DedPeter [7]

Answer:

$V=14373.33\ \text{mm}^3$

Explanation:

It is given that,

The radius of the inflated balloon is 7 cm.

We need to find the volume of air inside the balloon in milliliters.

The balloon is in the shape of a sphere whose volume is given by :

$V=\dfrac{4}{3}\pi r^3\\\\\text{Put r = 7 cm}\\\\V=\dfrac{4}{3}\times \dfrac{22}{7}\times 7\times 7\times 7\\\\V=1437.33\ \text{cm}^3\\\\\text{We know that, 1 cm = 10 mm}\\\\\text{So},\\\\V=14373.33\ \text{mm}^3$

Hence, the volume of the air inside the balloon is $14373.33\ \text{mm}^3$ .

3 0

2 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

3 years ago

The total number of the elements above with 2 valence electron is ____.

OverLord2011 [107]

Answer:

four elements

Explanation:

He = 1s²

Be = 1s² 2s²

C = 1s² 2s² 2p²

Mg = 1s² 2s² 2p⁶ 3s²

Si = 1s² 2s² 2p⁶ 3s² 3p²

Ca = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

Valence electrons are those electrons which are present in outer most electronic shell.

He, Be, Mg, Ca these four elements have 2 valence electrons.

While Si and C have four valance electrons.

4 0

2 years ago

6. 7. A hyperbaric chamber has a volume of 200. L. (a) How many moles of oxygen are needed to fill the chamber at room temperatu

Otrada [13]

Answer:

a) 24.7 mol

b) 790 g

Explanation:

Step 1: Given data

Volume of the chamber (V): 200. L

Room temperature (T): 23 °C

Pressure of the gas (P): 3.00 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 23°C + 273.15 = 296 K

Step 3: Calculate the moles (n) of oxygen

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 3.00 atm × 200. L/(0.0821 atm.L/mol.K) × 296 K = 24.7 mol

Step 4: Calculate the mass (m) corresponding to 24.7 moles of oxygen

The molar mass (M) of oxygen ga sis 32.00 g/mol. We will calculate the mass of oxygen using the following expression.

m = n × M

m = 24.7 mol × 32.00 g/mol = 790 g

6 0

3 years ago

Which is least likely to be reduced? <br> A. Zn^2+ <br> B. Fe^3+ <br> C. Cu^2+ <br> D. Fe^2+

alekssr [168]

A. Zn²⁺

<h3>Further explanation</h3>

Given

Cations of several elements

Required

The least to be reduced

Solution

If we look at the voltaic series:

The electrode which is easier to reduce than the hydrogen (H2) electrode has a positive sign (E red= +) and is located to the right of the voltaic series (right of H)

The electrode which is easier to oxidize than the hydrogen (H2) electrode and is difficult to experience reduction has a negative sign (E red= -) and is located to the left of the voltaic series (left of H)

Or you can look at the standard reduction potential value of the metals in the answer options, and the most negative reduction E° value which will be difficult to reduce.

The Zn metal is located far left of the other metals in the answer choices, so it is the most difficult to reduce

6 0

2 years ago