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3 years ago

Name the compound formed when a copper cation has a 2+ charge combined with a bromine anion that has a 1- charge.

Chemistry

1 answer:

Rashid [163]3 years ago

7 0

Answer: The compound is called copper (II) bromide or cupric bromide

Explanation:

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What is the ph of 0.001 m NaOH

julsineya [31]

Answer:

11.

Explanation:

For 0.001 M NaOH:

[OH⁻] = 0.001 M.

∵ pOH = - log[OH⁻]

∴ pOH = - log(0.001 M) = 3.

∴ pH = 14 - pOH = 14 - 3 = 11.

8 0

3 years ago

Under certain conditions,the solid and liquid states of water can exist in equilibrium. How are these conditions indicated on th

Oksi-84 [34.3K]

In a phase diagram that I have attached, the line marked in between ice and water (yellow and blue) is the conditions in which solid (ice) and liquid (Water) can coexist. As the graph shows, solid ice and liquid water coexist most when the temperature is zero, which should make sense since ice melts at 0, and water freezes as well at 0 degrees celsius.

3 0

4 years ago

What is the mass present in a 10.0L container of oxygen at a pressure of 105kPa and 20 degrees Celsius

omeli [17]

1.31 × 10⁴ grams.

<h3>Explanation</h3>

Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:

$P \cdot V = n \cdot R\cdot T$ ,

where

$P$ the pressure on the gas, $\bf P = 10^{5}\;\textbf{kPa}=10^{8}\;\textbf{Pa}$ ;
$V$ the volume of the gas, $V = 10.0 \;\text{L} = 10.0\times 10^{-3}\;\text{m}^{3}=10^{-2}\;\text{m}^{3}$ ;
$n$ the number of moles of the gas, which needs to be found;
$T$ the absolute temperature of the gas, $T=20\;\textdegree{}\text{C} = (20 + 273.15)\;\text{K} = 293.15\;\text{K}$ .
$R$ the ideal gas constant, $R = 8.314$ if P, V, and T are in their corresponding SI units: Pa, m³, and K.

Apply the ideal gas law to find $n$ :

$n = \dfrac{P\cdot V}{R\cdot T} = \dfrac{{\bf 10^{8}\;\textbf{Pa}}\times 10^{-2}\;\text{m}^{3}}{8.314 \;\text{Pa}\cdot\text{m}^{3}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times 293.15\;\text{K}} = 410.3\;\text{mol}$ .

In other words, there are 410.3 moles of O₂ molecules in that container.

There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be ${\bf 2}\times 16.00 = 32.00\;\text{g}$ . The mass of 410.3 moles of O₂ will be:

$410.3 \times 32.00 = 1.31\times10^{4}\;\text{g}$ .

What would be the mass of oxygen in the container if the pressure is approximately the same as STP at $10^{5}\;\textbf{Pa}$ or $10^{2}\;\text{kPa}$ instead?

6 0

4 years ago

How many grams of CO are needed to react with an excess of Fe2O3 to produce (210.3 g Fe) 209.7 g Fe? Show your work.

Fed [463]

Explanation:

$3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)$

1)Mass of CO when 210.3 g of Fe produced.

Number of moles of $Fe$ in 210.3 g=

$\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}$

$=\frac{210.3}{55.84 g/mol}=3.76 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : $\frac{3}{2}\times 3.76 moles$ of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

$\text{Number of moles}\times \text{molar mass of CO}=5.46\times 28 g/mol=157.92 g$

2)Mass of CO when 209.7 g of Fe produced.

Number of moles of $Fe$ in 209.7 g=

$\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}$

$=\frac{209.7}{55.84 g/mol}=3.75 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : $\frac{3}{2}\times 3.75 moles$ of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

$\text{Number of moles}\times \text{molar mass of CO}=5.625\times 28 g/mol=157.5 g$

4 0

3 years ago

Read 2 more answers

PLEASE NO BOTS lol

USPshnik [31]

The correct answer is

C. -432.4 J/mol K

8 0

3 years ago