Please help correct answer will get brainliest.

44. Convert the following. Express your answers in

professor190 [17]

Answer:

A = 0.75 ×10² KJ.

B = 3.9 ×10³ dg

C = 0.22 × 10² μl.

Explanation:

A = 7.5 ×10⁴ j to kilo joules

7.5 ×10⁴ / 1000 = 0.75 ×10² KJ.

Joule is the smaller unit while kilo joule is the larger unit. One kilo joule equals to the thousand joule that's why we will divide the given value by 1000 in order to convert into KJ.

B = 3.9 ×10⁵ mg to decigrams.

3.9 ×10⁵ / 100 = 3.9 ×10³ dg

Decigram is larger unit while milligram is smaller unit. One decigram is equal to the 100 milligram. In order to convert the given value into decigram we have to divide the value by 100.

C = 2.21 ×10⁻⁴ dL to micorliters

2.21 ×10⁻⁴ ×10⁵ = 0.22 × 10² μl.

Deciliter is bigger unit then micro liter . One deciliter equals to the 100000 micro liters. In order to convert the dL into micro liter we have to multiply the given value with 100000.

4 0

3 years ago

How do moss leaves and fish differ? How are they the same?

Lapatulllka [165]

Moss leaves and fish are different in that, the moss leave is a producer, that is, it produces its own food through photosynthesis while the fish is a consumer, it feeds on foods that are not produced by it.
Both moss and fish are the same in the sense that both have cell as their basic unit of life, that is, they both possess cells.

4 0

3 years ago

What observations can you make about this group of people?

DedPeter [7]

Answer:The people are part of a softball team. They appear happy, they are showing number one with their fingers, and they are holding a trophy. They probably won the softball league championship

Explanation:

7 0

3 years ago

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

3 years ago

2 CuCl2 +4KI -> 2 Cul + 4 KCI + 12

Amanda [17]

Answer:

Explanation:

This is an example of a limiting reactant question, and is very common as a general chemistry problem.

We first see the balanced equation, that is:

2CuCl2+4KI→2CuI+4KCl+I2

We first need to find the limiting reactant

We see that 0.56 g of copper(II) chloride (CuCl2) reacts with 0.64 g of potassium iodide (KI) . So, let's convert those amounts into moles.

Copper(II) chloride has a molar mass of

134.45 g/mol . So in 0.56 g of copper(II) chloride, then there exist

0.56g134.45g/mol≈4.17⋅10−3 mol

Potassium iodide has a molar mass of

166 g/mol . So, in 0.64 g of potassium iodide, there exist

if it wrong i am sorry

5 0

3 years ago