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4 years ago

Warming hand over a radiator is called?

Chemistry

2 answers:

allochka39001 [22]4 years ago

5 0

Warming hand over a radiator is called radiation

Digiron [165]4 years ago

3 0

A warming hand over a radiator is called a Radiation

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What two factors must be held constant for density to be a constant ratio?

Neko [114]

Answer:

Temperature and Pressure

Explanation:

Temperature and pressure cause change in volume.

So any change in volume will alter the ratio of density as given by equation of density.

Density = mass/ volume

Change in volume will alter the ratio.

Kindly mark it branliest if the answer is little bit satisfying.

7 0

3 years ago

Carbon disulfide burns in oxygen to yield carbon dioxide and sulfur dioxide according to the following chemical equation. CS2(l)

Semmy [17]

Answer:

a) the limiting reactant is 02

b) There will remain 0.667 moles of CS2

c) There will be formed 0.333 moles oof CO2 and 0.667 moles of SO2

Explanation:

Step 1: Data given

Number of moles of CS2 = 1.00 mol

Number of moles of O2 = 1.00 mol

Molar mass of O2 = 32 g/mol

Molar mass of CS2 = 76.14 g/mol

Step 2: The balanced equation

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

Step 3: Calculate the limiting reactant

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

O2 is the limiting reactant. It will completely be consumed.(1.00 mol).

CS2 is in excess. There will react 1.00/ 3 = 0.333 moles

There will remain 1.00 - 0.333 = 0.667 moles of CS2

Step 4: Calculate moles of CO2 and SO2

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

For 1.00 mol of O2 we have 1.00/3 = 0.333 moles CO2

For 1.00 mol of O2 we have 1.00 /(3/2) = 0.667 moles of SO2

8 0

3 years ago

How many moles are in 25 grams of HF

sergey [27]

Using this equation, we can take 25/(1.0 + 19) and find that it is equal to 1.25 moles.

5 0

3 years ago

Read 2 more answers

Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+

DIA [1.3K]

These are four questons and four answers:

Answers:

1) 7⁺

2) 1⁻

3) 3⁺

4) 5⁺

Explanation:

Question 1) Cl₂O₇:

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

2×X + 7 (-2) = 0

2X - 14 = 0

2X = +14

X = +14 /2 = + 7

Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.

Question 2) AlCl₄⁻

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

1 (+3) + 4X = - 1

+3 + 4X = - 1

4X = - 1 - 3

4X = - 4

X = - 1

Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.

Question 3) Ba(ClO₂)₂

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

+2 + 2X + 4 (-2) = 0

2X +2 - 8 = 0

2X - 6 = 0

2X = +6

X = + 3

Conclusion: the oxidation number of Cl in Ba(ClO₂)₂ is 3⁺.

Question 4) CIF₄⁺

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

X + 4(-1) = +1

X - 4 = +1

X = +1 + 4

X = + 5

Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.

6 0

3 years ago

Consider a sample of calcium carbonate in the form of a cube measuring 2.805 in. on each edge.

Naya [18.7K]

Answer:

The answer to your question is: 8.82 x 10 ²⁴ atoms of oxygen

Explanation:

Data

Cube measuring : 2.805in on each edge

density = 2.7 g/cm3

# of oxygen atoms in the cube = ?

Process

Volume of the cube = (2.805)³ = 22.07 in³

Convert in³ to cm³ 1 in³ ------------------- 16.39 cm³

22.07 in³ -------------------- x

x = (22.07 x 16.39) / 1 = 361.72 cm³

Density = mass / volume

Mass = density x volume

Mass = (2.7)(361.72) = 976.66 g

Atomic mass CaCO3 = 40 + 12 + (16 x 3) = 100 g

100 g of CaCO3 -------------------- 48 g of O2

976.66 g of CaCO3 --------------- x

x = (976.66 x 48) / 100 = 468.8 g of O2

32 g of O2 ---------------------- 1 mol

468.8 g of O2 -------------------- x

x = (468.8 x 1) / 32 = 14.65 mol

1 mol ----------------------- 6.023 x 10 ²³ atoms

14.65 mol ------------------- x

x = (14.65 x 6.023 x 10 ²³) / 1 = 8.82 x 10 ²⁴ atoms of oxygen

3 0

3 years ago