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3 years ago

Is the shape of solid defined or undefined

Chemistry

2 answers:

enyata [817]3 years ago

6 0

defined, a solid will always have a definite shape and volume because it doesn´t depend on its container shape

Fittoniya [83]3 years ago

4 0

A solid has a definite shape

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Two atoms with the same number of protons but a different number of neutrons are

Luda [366]

They are called isotopes.

Example of isotopes are Hydrogen and deuterium.
Hydrogen is 1 proton and 0 neutrons.
Deuterium is 1 proton and 1 neutron

6 0

3 years ago

. How many grams of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide? ____Cl

dlinn [17]

Answer: 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide.

Explanation:

The balanced chemical reaction is;

$Cl_2+MgBr_2\rightarrow MgCl_2+Br_2$

$Cl_2$ is the limiting reagent as it limits the formation of product and $MgBr_2$ is the excess reagent.

According to stoichiometry :

1 mole of $Cl_2$ produces = 1 mole of $MgCl_2$

Thus 2 moles of $Cl_2$ will produce= $\frac{1}{1}\times 2=2moles$ of $MgCl_2$

Mass of $MgCl_2=moles\times {\text {Molar mass}}=2moles\times 95g/mol=190g$

Thus 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide

5 0

2 years ago

John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det

navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

Answer the question below based on the periodic table entry for bromine.

Shalnov [3]

The number of neutrons bromine will have are equal to
= protons + neutrons
so,
80-35=45

4 0

3 years ago

Read 2 more answers