Compound A solid
Compound B solid
Compound
C
Compound
D
B
C
gas
D
solid
808
1610
-56
680
yes
no
no
Compound B because it has a very high melting point.
yes
You are studying the differences between ionic and covalent compounds. After studying the properties of each
type of compound, your teacher has provided this table and asked you to (1) identify the ionic compound(s) and
explain your choice(s).
A
Compounds A and D because they have relatively high melting points.
AROLIChessies
Compounds A, B, and C because they are solid at room temperature.
Compounds A and D because the conduct electricity in aqueous solutions.
yes
no
yes
yes

Answer:

La respuesta correcta es: Al aumentar la temperatura se dilata con facilidad

Explanation:

El mercurio (Hg) es un elemento metálico que a temperatura ambiente es un líquido. Tiene la propiedad de dilatarse con facilidad al aumentar la temperatura, propiedad que ha sido utilizada en los termómetros de mercurio. En estos termómetros, el mercurio está dentro de un tubo de vidrio graduado con un bulbo para transmitir el calor. Al estar en contacto con un cuerpo más caliente (a mayor temperatura), el mercurio del interior del termómetro se dilata (expande su volumen) y asciende por el tubo indicando la temperatura correspondiente, graduada en el vidrio.

Answer:

iron (III) oxide is a gas

Answer:

0.065 meters

Explanation:

Answer:

Start by dividing the length value by 1000, so that would be

65 / 1000 = 0.065 meters

Explanation:

A-Corrosion

B- Magnesium helps ur body regulates its zinc levels and zinc absorbs into absorbs magnesium more efficiently

Answer:

false

Explanation:

trust

Answer:

True

Explanation:

Because they require different things and have different functions

Answer:

3

Explanation:

The plates are moving away from each other

Answer:

Atacama

Explanation:

Answer:

made of different kinds of tissues

Explanation:

cause that's the only one that makes sense because it goes: cell, tissues,organ,organ system , organism

1. the simplest unit of life - cell

2. a individual organ - would be made of different kinds of tissues

3. tissues - would be made of one kind of cell

4- an organ system - meets all the needs of a organism

also it wouldn't make sense for a individual organ to be made of only one kind of cell so yeah

good luck :) I hope this was right

I think the answer is c if not then try a

Answer: B

Explanation:

Answer:

D

Explanation:

31 / 2.8 = 11.0714286 L per mole of helium

3.5 / 4 = 0.875 moles

2.8 + 0.875 = 3.675 moles

11.0714286 x 3.675 = 40.6875 L

Answer:

Both graduated cylinders and beakers are pieces of laboratory glassware that have a specific function. Graduated cylinders typically are more accurate at reading the volumes of the liquid inside. Beakers are better for stirring and mixing liquids.

Flasks are notable for their unique shape: a rounded vessel and a cylindrical neck. The main differing characteristic between a flask and a beaker is that beakers have straight sides, rather than slanted sides like a flask. Beakers are mainly for measuring and transporting liquids from one site to the next.

Explanation:

hope this helps!!!!

Answer:

2.41 mol

Explanation:

53.9 L x 1 mol/22.4 L = 2.41 mol

Answer:

2.7 mL

Explanation:

The equation of the reaction is;

2AlBr3 + 3Cl2 -----> 2AlCl3 + 3Br2

Number of moles in 2.12 x 1022 formula units of aluminum bromide

1 mole of AlBr3 = 6.02 * 10^23 formula units

x moles = 2.12 x 1022 formula units

x = 2.12 x 1022 formula units  * 1 mole/ 6.02 * 10^23 formula units

x = 0.0352 moles of AlBr3

According to the reaction equation;

2 moles of AlBr3 produces 3 moles of Br2

0.0352 moles of AlBr3 produces 0.0352 moles  *  3 moles /2 moles

= 0.0528 moles of Br2

Mass of Br2 produced = 0.0528 moles of Br2 * 159.808 g/mol

Mass of Br2 produced = 8.44g

But density = mass/volume

volume = mass/density

volume of Br2 = 8.44 g/ 3.12 g/mL

volume of Br2 =  2.7 mL

6, 0

thats the answer, i hope it helps

Answer:

8.334×10^-6 moles

Explanation:

a mole is 6.023×10^23

5.02×10^18 ÷ 6.023×10^23=8.33×10^-6

so there are 8.33×10^-6 moles of water in 5.02x10^18 molecules of water

The number of moles of water in which 5.02 × 10¹⁸ molecules of water is present is 8.33 × 10⁻⁶.

Moles is a unit which is used to tells about the quantity of any substance, in one moles of any substance 6.023 × 10²³ number of atoms of that substances are present.

In 1 mole of water = 6.023 × 10²³ number of molecules of water is present

So, number of 5.02 × 10¹⁸ molecules of water present in how many moles will be calculated as:
Required moles = 5.02 × 10¹⁸ / 6.023 × 10²³ = 8.33 × 10⁻⁶

Hence, required moles is 8.33 × 10⁻⁶.

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Answer:

HCI is a strong acid : HCI -> h^+  + CI^-                                                              [ H^+ ] = 1.0  x 10 ^-3  -> pH = -log[ H^+ ] = -log 10 ^-3 = 3

Explanation:

Answer:

0.95g of Ag we can produce

Explanation:

To solve this question we have to convert the mass of silver nitrate (AgNO₃) to moles. The moles of silver nitrate = Moles of Ag. Then, with molar mass of Ag we can know the mass of silver:

Moles AgNO₃ -Molar mass: 169.87g/mol-:

1.5g * (1mol / 169.87g) = 0.0088 moles Ag

Mass Ag -Molar mass: 107.87g/mol-:

0.0088 moles Ag * (107.87g / mol) =

Genetic engineering improved the lives if individuals with insulin because it helps the body use and store sugar, since people with diabetes can't produce insulin.

Answer:

c) %5.68×10²⁴

Explanation:

Number atoms=Avogadro's number ₓ number of moles

N= 6.023ₓ 10²³×9.44

N=5.68×10²⁴

Answer:

Mass of 1 mole of Aluminium = Mass of 6.02 x 10 23atoms of Aluminium =  26.98 grams.

Explanation:

1 mole of any element or molecule has 6.02 x 10 23atoms or subparts.

Hence 1 mole of aluminum will also have 6.02 x 10 23atoms. Hence, it is clear that mass of 1 mole of Aluminium will be equal to 6.02 x 10 23atoms of Aluminium.

Thus,

Mass of 1 mole of Aluminium = Mass of 6.02 x 10 23atoms of Aluminium =  26.98 grams.

Solution :

For the real gas following Vander Waals equation, reversible isothermal work done is given by :

[tex]$W= \int P \ dV$[/tex]

[tex]$W= \int \left(\frac{nR}{V-nb}-\frac{an^2}{V^2}\right) dV$[/tex]

[tex]$W_{real}=-nRT \ln \left[\frac{V_2-b}{V_1-b}\right]-an^2\left[\frac{1}{V_2}-\frac{1}{V_1}\right]$[/tex]

Given : [tex]$V_1=0.1 \ dm^3$[/tex], [tex]$$V_2 =100 \ dm^3[/tex]  

[tex]$a=359 \ atm \ L^2/ mol^2, \ \ b = 0.0427 \ L/mol$[/tex]

As T is not given, assuming T = 290 K

[tex]$W_{real} = 0.0823 \times 240 \left[ \ln \left(\frac{100-0.042}{0.1-0.0427}\right)\right]-3.59\left[\frac{1}{100}-\frac{1}{0.1}\right]$[/tex]

[tex]$W_{real}= -183.06+35.864$[/tex]

[tex]$W_{real}=-147.196$[/tex]

For the perfect gas,

[tex]$W=nRT \ln\left[\frac{V_2}{V_1}\right] =(1)(0.0823)(290) \ln \left[\frac{100}{0.1}\right]$[/tex]

[tex]$W_{ideal}=169.415$[/tex] kJ

Answer:

c) 7.2 moles

Explanation:

7.2 moles of [tex]CO_2[/tex]  are produced during the complete combustion of 3.6 moles of [tex]C_2H_6[/tex].  Hence, option C is correct.

A mole is defined as 6.02214076 ×[tex]10^{23}[/tex]

of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Balance your chemical reaction first.

[tex]2C_2H_6 + 7O_2[/tex] —> [tex]4CO_2 + 6H_2O[/tex]

2 moles of [tex]C_2H_6[/tex] produce 4 moles of [tex]CO_2[/tex].

So, 3.6  moles of [tex]C_2H_6[/tex]  produce= (4÷2)x3.6 =7.2 moles

Hence, 7.2 moles of [tex]CO_2[/tex]  are produced during the complete combustion of 3.6 moles of [tex]C_2H_6[/tex].

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Answer:

The answer is B

Explanation:

I'm just big-brained

And because it's hotter than the rest and since all of them are cold "B" is just going to make them warmer.

Answer:

KCl[tex]_{2}[/tex], >  KBr[tex]_{2}[/tex], > KI[tex]_{2}[/tex]

Explanation:

KBr[tex]_{2}[/tex],   KCl[tex]_{2}[/tex],   KI[tex]_{2}[/tex]

The equilibrium constant for the reaction between Alkene and Halogen to give Alkyl halide as a product

Keq = [tex]\frac{[C2H4X2 ]}{[C2H4 ][ X2]}[/tex]

hence the order of equilibrium from the Largest to the smallest equilibrium

= KCl[tex]_{2}[/tex], >  KBr[tex]_{2}[/tex], > KI[tex]_{2}[/tex]