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2 years ago

2. Which process is occurring in this photograph of a glacier?

Chemistry

2 answers:

nirvana33 [79]2 years ago

4 0

Answer:

The answer is D. have a good day!

Explanation:

Paul [167]2 years ago

4 0

Answer:

D. Plucking

Explanation:

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katrin [286]

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3 years ago

List first thirty elements with their valences

lbvjy [14]

Element Atomic Number Valency

Valency of Hydrogen 1 1

Valency of Helium 2 0

Valency of Lithium 3 1

Valency of Beryllium 4 2

Valency of Boron 5 3

Valency of Carbon 6 4

Valency of Nitrogen 7 3

Valency of Oxygen 8 2

Valency of Fluorine 9 1

Valency of Neon 10 0

Valency of Sodium (Na) 11 1

Valency of Magnesium (Mg) 12 2

Valency of Aluminium 13 3

Valency of Silicon 14 4

Valency of Phosphorus 15 3

Valency of Sulphur 16 2

Valency of Chlorine 17 1

Valency of Argon 18 0

Valency of Potassium (K) 19 1

Valency of Calcium 20 2

Valency of Scandium 21 3

Valency of Titanium 22 4

Valency of Vanadium 23 5,4

Valency of Chromium 24 2

Valency of Manganese 25 7, 4, 2

Valency of Iron (Fe) 26 2, 3

Valency of Cobalt 27 3, 2

Valency of Nickel 28 2

Valency of Copper (Cu) 29 2, 1

Valency of Zinc 30 2

5 0

2 years ago

How many acetyl coa molecules may be obtained from oxidation of an 18-carbon fatty acid?

saveliy_v [14]

The answer would be 9

7 0

3 years ago

Consider the following chemical reaction:

laiz [17]

Answer:

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol

Step 3: Calculate the moles of SO₃ produced

0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃

Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

0.0736 mol × 22.4 L/1 mol = 1.65 L

6 0

2 years ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

3 years ago