CHEMISTRY COLLEGE

Find the density of an object that has a volume of 2.3 x 10^2 and a mass of
3.5 x 10^3. *

Answers

Answer 1

Answer:

Answer:

$(250)/(23)$

Step-by-step explanation:

$10^(2) = 100$
2.3 × 100 = 230
$10^(3) = 1000$
3.5 × 1000 = 3500
$D = (m)/(v)$
$D = (3500)/(230)$
3500 ÷ 230 = $(250)/(23)$

I hope this helps!

Related Questions

Answer the question please.
Increasing which factor will cause the gravitational force between two objects to decrease?weights of the objectsdistance between the objectsacceleration of the objectsmasses of the objects
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.
How many water molecules are in a block of ice containing 1.50 mol of water (H2O)?
Be sure to answer all parts. A baseball pitcher's fastballs have been clocked at about 97 mph (1 mile = 1609 m). (a) Calculate the wavelength of a 0.148−kg baseball (in nm) at this speed. × 10 nm (Enter your answer in scientific notation) (b) What is the wavelength of a hydrogen atom at the same speed? nm

Oxygen is a __________ and nitrogen is a __________. metalloid, metalloid nonmetal, metal nonmetal, nonmetal nonmetal, metalloid metal, metalloid

Answers

Answer:

"nonmetal, nonmetal"

Explanation:

Oxygen is a non metal and Nitrogen is a non metal. It is 8th element of the periodic table. It is located in period 2 and group 16.

Nitrogen lies at the group 15 of the periodic table. Its atomic no is 7. Its valency is 2.

Hence, the correct option is (c) "nonmetal, nonmetal".

Final answer:

Oxygen and nitrogen are both nonmetals. They are unable to conduct heat or electricity effectively and are typically found on the right side of the periodic table.

Explanation:

In the periodic table of elements, oxygen and nitrogen are both classified as nonmetals. Nonmetals are elements that are not able to conduct electricity or heat very well. As opposed to metals, nonmetals are brittle and do not have the ability to be shaped into thin sheets or wires. They are typically found on the right side of the periodic table and are represented by groups 14-17. So, to answer the student's question, oxygen is a nonmetal and nitrogen is a nonmetal.

Learn more about Nonmetals here:

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How many joules of heat energy are absorbed when 80.0 g of water are heated from 10.0°C to 50.0°C? *

Answers

Answer:

13440 J

Explanation:

c ≈ 4200 J / (kg * °C)

m = 80 g = 0,08 kg

$t_(1)$ = 10 °C

$t_(2)$ = 50 °C

The formula is: Q = c * m * ( $t_(2) - t_(1)$ )

Calculating:

Q = 4200 * 0,08 * (50 - 10) = 13440 (J)

HELP!!!!!!
I DON'T KNOW THE ORDER!!

Answers

Answer:

for 1 solid its freezing.

for 2 solid and liquid its melting

for 6 liquid to gas its evaporation and for 5 gas to liquid its condensation.

Explanation:

hope this helped :)

Answer:

solid->liquid= melting

liquid->solid= freezing

gas->liquid= consendation

liquid->gas= evaporation

Calculate the solubility of CuX (Ksp=[Cu2+][X2−]=1.27×10−36) in a solution that is 0.200 M in NaCN.I have already tried to square root the Ksp value to get the answer but it was wrong.

Answers

Answer:

Solubility= 1.08×10-12

Explanation:

Take the cube root of 1.27×10-36

Answer:

The solubility of CuX is 1.425x10⁻⁷M

Explanation:

Given:

initial concentration of NaCN=0.2M

Ksp=1.27x10⁻³⁶

The reaction are:

CuX → Cu²⁺ + X²⁺, Ksp=1.27x10⁻³⁶

Cu²⁺ + 4CN⁻ → (Cu(CN)₄)²⁻, Kf=1x10²⁵

The overall reaction is:

CuX + 4CN⁻ → (Cu(CN)₄)²⁻ + X²⁺

The equilibrium constant is:

K=Ksp*Kf=1.27x10⁻³⁶*1x10²⁵=1.27x10⁻¹¹

CuX + 4CN⁻ → (Cu(CN)₄)²⁻ + X²⁺

I - 0.2 0 0

C - -4 +x +x

E - 0.2-4 x x

The equation for equilibrium is:

$K=([Cu(CN)4]^(2) [X])/([CN]^(4) ) \n1.27x10^(-11) =(x^(2) )/((0.2-x)^(4) )$

Here, solving for x:

x=1.425x10⁻⁷M=CuX

How many s’mores can you make from the following combinations? What is the limiting reagent?

Answers

The limiting reagent and the number of S'mores produced for each of the reactions is given below:

Reaction 1. The limiting reagent is Cp; 1.6 S'mores are produced.

Reaction 2. The limiting reagent is M;2 S'mores are produced.

Reaction 3. The limiting reagent is Gc; 2.5 S'mores are produced.

Reaction 4. The limiting reagent is M;1 S'more is produced.

Stoichoimetry

Stoichiometry is the process of measuring quantitatively the mass and quantity relationships among reactants and products in a given reaction.

The equation of the reaction shows the stoichiometry between reactants and products.

For the given reaction, the equation of reaction is as follows:

1M + 2Gc + 3Cp ----> 1Gc2MCp3

where:

Gc = Graham Cracker
M = Marshmallow
Cp = Chocolate pieces
S’more = Gc2MCp3

From the equation of reaction:

1 marshmallow, 2 Graham cracker and 3 chocolate pieces are required to make 1 S'more

Calculating the number of S'mores and the limiting reactant of the given reaction

The stoichiometric equation is: 2Gc + 1M + 3Cp ----> 1Gc2MCp3

The ratio of Gc to M to Cp is 2 : 1 : 3

Reaction 1. 4 Gc + 2M + 5 Cp

The ratio of Gc to M to Cp in the reaction above is 2 : 1 : 2.5

Therefore the limiting reagent is Cp

3 Cp makes 1 S'more

5 Cp will make 5 * 1/3 S'more = 1.6 S'mores

Therefore, 1.6 S'mores are produced.

Reaction 2. 6 Gc + 2M + 9 Cp

The ratio of Gc to M to Cp in the reaction above is 3 : 1 : 4.5

Therefore the limiting reagent is M

1 M makes 1 S'more

2 Cp will make 2 * 1/1 S'more = 2 S'mores

Therefore, 2 S'mores are produced.

Reaction 3. 5 Gc + 3M + 9 Cp

The ratio of Gc to M to Cp in the reaction above is 1.6 : 1 : 3

Therefore the limiting reagent is Gc

2 Gc makes 1 S'more

5 Gc will make 5 * 1/2 S'more = 2.5 S'mores

Therefore, 2.5 S'mores are produced.

Reaction 4. 7 Gc + 1M + 6 Cp

The ratio of Gc to M to Cp in the reaction above is 7 : 1 : 6

Therefore the limiting reagent is M

1 M makes 1 S'more

Therefore, 1 S'more is produced.

The limiting reagent and the number of S'mores produced for each of the reactions is given below:

1. The limiting reagent is Cp; 1.6 S'mores are produced.

2. The limiting reagent is M;2 S'mores are produced.

3. The limiting reagent is Gc; 2.5 S'mores are produced.

4. The limiting reagent is M;1 S'more is produced.

Learn more about Stoichiometry and limiting reagents at: brainly.com/question/14222359

What is the chemical formula for the ionic compound formed by Au3+ and
HSO3-?

Answers

Answer:

The chemical formula for the ionic compound formed by Au3+ and

HSO3-compound is Au(HSO3)3

Explanation:

The charge on Au ion is $+3$

And the charge on HSO3- is $-1$

Thus, the number of atoms required by HSO3- to complete its octate is 1. On the other hand Au has 3 excess ions and hence it is to be released to reach the stable state.

So three molecules of HSO3- will combine with one atom of Au 3+

Thus, the compound formed by these two is Au(HSO3)3

Final answer:

The chemical formula for the ionic compound formed by Au3+ and HSO3- is Au(HSO3)3, as ionic compounds are always neutral.

Explanation:

The ionic compound formed by Au3+ (Gold ion) and HSO3- (Bisulfite ion) must have a net charge of zero since ionic compounds are neutral. Hence, we need 3 bisulfite ions to balance out one gold ion, which gives us the chemical formula as Au(HSO3)3.

Indeed, the formation of ionic compounds is a fascinating process. It involves the transfer of electrons from one atom (usually a metal) to another (usually a nonmetal), resulting in the formation of ions. These ions are then attracted to each other due to their opposite charges, forming an ionic compound. In this case, the gold ion (Au3+) donates three electrons, which are accepted by three bisulfite ions (HSO3-). This results in a neutral compound, as the positive and negative charges balance each other out. The resulting compound, Au(HSO3)3, is an example of how elements can combine in specific ratios to form neutral compounds.