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3 years ago

What is the name of this compound XeF4?

1 answer:

5 0

Chemical nomenclature, replete as it is withcompounds with complex names, is arepository for some very peculiar and sometimes startling names. A browse through the Physical Constants of Organic Compounds in the CRC Handbook of Chemistry and Physics (a fundamental resource) will reveal not just the whimsical work of chemists, but the sometimes peculiar compound names that occur as the consequence of simple juxtaposition. Some names derive legitimately from their chemical makeup, from the geographic region where they may be found, the plant or animal species from which they are isolated or the name of the discoverer.

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The rate of filtration is Select one: a. modified by boundary layer effects. b. increased with increasing viscosity of the fluid

brilliants [131]

Answer:

c. decreased with increasing viscosity of the fluid being filtered.

Explanation:

Filtration is a method of separation which is useful in separating solids from the fluids (either liquids or gases) by adding a specific medium through which only fluid can pass. Fluid which passes through the filter is known as the filtrate.

Increasing the viscosity of the fluid which is being filtered decreases the rate of filtration as the liquid being more viscous take more time to cross the pores of the filter. Thus, rate of filtration is indirectly proportional to the viscosity of the fluid which is being filtered.

So, option C is correct.

3 0

2 years ago

How do microorganisms influence the chemical characteristics of soil? Explain Please.

galben [10]

Answer:

Soil microbes are also important for the development of healthy soil structure. ... Soil microorganisms are both components and producers of soil organic carbon, a substance that locks carbon into the soil for long periods. Abundant soil organic carbon improves soil fertility and water-retaining capacity.

Explanation:

7 0

2 years ago

a sample of the hydrate of sodium carbonate has a mass of 8.85 g. it loses 1.28 g when heated. find the molar ratio of this comp

xenn [34]

Answer:

Molar ratio of the compound is 1:1 and the type of hydrate is Mono hydrate.

Explanation:

From the given,

Mass of sodium carbonate $Na_{2}CO_{3}.XH_{2}O$ = 8.85 g

Loss mass $H_{2}O$ = 1.28 g

Actual weight of sodium carbonate = 8.85 g - 1.28 g = 7.57 g

$7.57 g Na_{2}CO_{3} \times \frac{1mol}{106 g} =\frac{0.0714}{0.0714}=1$

$1.28g H_{2}O \times \frac{1mol}{18 g} =\frac{0.0711}{0.0714}=1$

Therefore, the compound has only one water molecule.

Molecular formula of the compound is $Na_{2}CO_{3}.H_{2}O$ an name of the compound is sodium carbonate mono hydrate.

Hence, the type of the compound is Mono hydrate.

8 0

2 years ago

Write a balanced net ionic equation for the following double replacement reaction

shtirl [24]

Answer:

Ba²⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s)

Explanation:

There are three steps you must follow. You must write the:

Molecular equation
Ionic equation
Net ionic equation

A. Molecular equation

BaCl₂(aq) + Na₂SO₄(aq) ⟶ BaSO₄(s) + 2NaCl(aq)

B. Ionic equation

You write all the soluble substances as ions.

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

C. Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

The net ionic equation is

Ba²⁺(aq) + SO₄²⁻(aq) ⟶ BaSO₄(s)

3 0

3 years ago

Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?

WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

6 0

2 years ago