What type of moon cycle do solar and lunar Eclipse occurs in

One sphere has a radions of 181 cm, another has a radius of 5.01 cm. What is the difference in volume (in cubic centimeters) bet

PSYCHO15rus [73]

Answer:

$2.4*10^{7}cm^{3}$

Explanation:

First you should calculate the volume of a big sphere,so:

$V_{big}=\frac{4}{3}\pi r^{3}$

$V_{big}=\frac{4}{3}\pi (181cm)^{3}$

$V_{big}=2.4*10^{7}cm^{3}$

Then you calculate the volume of a small spehre, so:

$V_{small}=\frac{4}{3}\pi r^{3}$

$V_{small}=\frac{4}{3}\pi (5.01cm)^{3}$

$V_{small}=5.3*10^{2}cm^{3}$

Finally you subtract the two quantities:

$V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}$

$V_{big}-V_{small}=2.4*10^{7}cm^{3}$

5 0

3 years ago

What are the products of the neutralization reaction between HCl and

liraira [26]

Answer:

opinion b

Explanation:

the product of neutralization reaction between hcl and CA(oh)2 is option b.

4 0

2 years ago

The ion n3− has _____ protons and _____ electrons.enter your answers as integers separated by a comma.

bearhunter [10]

The ion N³⁻ is called the azide ion. In its neutral state, it occurs as the element Nitrogen. The atomic number of Nitrogen is 7. When it turns into an anion (negatively charged ion), it gains 3 more electrons. That's why its net charge becomes -3. It means that the protons is still 7, but the electrons are now 10.

Overall charge = +7 + -10 = -3

5 0

3 years ago

Read 2 more answers

What is the volume of 40.0 grams of argon gas at STP ?

MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Aqueous Solutions

States of Matter

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 40.0 g Ar

[Solve] L Ar

Step 2: Identify Conversions

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

Step 3: Convert

[DA] Set up: $\displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 24.9235 \ L \ Ar$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

24.9235 L Ar ≈ 24.9 L Ar

5 0

3 years ago

A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to

melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0

3 years ago