During an experiment water was heated in a container covered with a glass lid. The following observations were recorded.

How do scientist classify newly discovered organisms into domains and kingdoms? (6th Grade)

olga55 [171]

Answer: Scientists look for cell types, the ability to make food, number of cells in their bodies.

Explanation: Scientists look for cell types, the ability to make food, number of cells in their bodies when placing newly-discovered organisms.

5 0

3 years ago

Vitamin K is involved in normal blood clotting. When 0.802 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point

xenn [34]

Answer:

Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]

Explanation:

The freezing point of camphor = 178.4 ⁰C

the Kf of camphor = 37.7°C/m

where : m = molality

the relation between freezing point depression and molality is

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant of camphor

molality = moles of solute dissolved per kg of solvent.

putting values

2.69°C = 37.7°C/m X molality

molality = 0.0714 mol /kg

$molality=\frac{molesofvitaminK}{massofcamphor(kg)}=\frac{moles}{0.025}$

moles of vitamin K = 0.0714X0.025 = 0.00178 mol

we know that moles are related to mass and molar mass of a substance as:

$moles=\frac{mass}{molarmass}$

For vitamin K the mass is given = 0.802 grams

therefore molar mass = $\frac{mass}{moles}=\frac{0.802}{0.00178}=450.56\frac{g}{mol}$

7 0

3 years ago

Write the Formulas for these ionic compounds:

Alinara [238K]

Answer:

<h2>1. Tin(IV) oxide</h2>

Tin(IV) oxide, also known as stannic oxide, is the inorganic compound with the formula SnO₂. The mineral form of SnO₂ is called cassiterite, and this is the main ore of tin. With many other names, this oxide of tin is an important material in tin chemistry. It is a colourless, diamagnetic, amphoteric solid.

Formula: SnO₂

Molar mass: 150.71 g/mol

Melting point: 1,630 °C

Density: 6.95 g/cm³

<h2>2. Aluminum oxide</h2>

Aluminium oxide is a chemical compound of aluminium and oxygen with the chemical formula Al₂O₃. It is the most commonly occurring of several aluminium oxides, and specifically identified as aluminium(III) oxide.

Formula: Al₂O₃

Molar mass: 101.96 g/mol

Melting point: 2,072 °C

Density: 3.95 g/cm³

Boiling point: 2,977 °C

ChemSpider ID: 14086

ChemSpider ID: 26988

<h2>3. Silver sulfide</h2>

Silver sulfide is an inorganic compound with the formula Ag ₂S. A dense black solid, it is the only sulfide of silver. It is useful as a photosensitizer in photography. It constitutes the tarnish that forms over time on silverware and other silver objects.

Formula: Ag2S

Molar mass: 247.8 g/mol

Density: 7.23 g/cm³

ChemSpider ID: 145878

<h2>4. Copper (II) sulfate</h2>

Copper(II) sulfate, also known as copper sulphate, are the inorganic compounds with the chemical formula CuSO₄ₓ, where x can range from 0 to 5. The pentahydrate is the most common form. Older names for this compound include blue vitriol, bluestone, vitriol of copper, and Roman vitriol.

Formula: CuSO4

IUPAC ID: Copper(II) sulfate

Molar mass: 159.609 g/mol

Density: 3.6 g/cm³

Melting point: 110 °C

Soluble in: Water

<h2>5. Potassium carbonate</h2>

Potassium carbonate is the inorganic compound with the formula K₂CO₃. It is a white salt, which is soluble in water. It is deliquescent, often appearing as a damp or wet solid. Potassium carbonate is mainly used in the production of soap and glass.

Formula: K2CO3

Molar mass: 138.205 g/mol

IUPAC ID: Potassium carbonate

Density: 2.43 g/cm³

Melting point: 891 °C

Soluble in: Water

Explanation:

Hope it is helpful....

4 0

3 years ago

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl),

choli [55]

Answer:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

B. 100.00° C

C. 101.02° C

D. 100.51° C

Which of the following solutions will have the lowest freezing point?

A. 1.0 mol/kg sucrose (C12H22O11)

B. 1.0 mol/kg lithium chloride (LiCl)

C. 1.0 mol/kg sodium phosphide (Na3P)

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

B. magnesium sulfate (MgSO4)

C. potassium bromide (KBr)

D. All will be equally effective.

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

A. 98.7 kPa

B. 96.3 kPa

C. 101.3 kPa

D. 100.2 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

A. 101.2°C

B. 105.9°C

C. 102.7°C

D. 108.1°C

Explanation:

The following are the answers to the different questions:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

Which of the following solutions will have the lowest freezing point?

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

B. 96.3 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

D. 108.1°C

5 0

3 years ago

When 1.0g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (C4H10O, Hvap= 15.7kj/mo

Aloiza [94]

Answer:

Option D = 0.2 Kj

Explanation:

Given data:

Mass of diethyl ether = 1.0 g

Hvap = 15.7 Kj / mol

Heat absorbed = ?

Solution:

Q = mass × Hvap / molar mass

Q = 1.0 g × 15.7 Kj / mol / 74.12 g/mol

Q = 15.7 Kj / 74.12

Q = 0.212 KJ

8 0

3 years ago