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BaLLatris [955]
2 years ago
13

Describe some of the dangers of Global warming.

Chemistry
1 answer:
grin007 [14]2 years ago
8 0

Answer:

Increased Heat, Drought, and Insect outbreaks.

Explanation:

Increased heat, drought and insect outbreaks, all linked to climate change, have increased wildfires. Declining water supplies, reduced agricultural yields, health impacts in cities due to heat, and flooding and erosion in coastal areas are additional concerns.

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Which nitrogenous base is NOT found in DNA? *
scoundrel [369]

Answer:

c. uracil

Explanation:

uracil is not found in DNA. the missing base would be thymine

5 0
3 years ago
Please help as soon as possible please and thank you.
HACTEHA [7]

Answer:

The answer is A. I took the test today.

Explanation:

8 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
2 years ago
A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water
Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be (C_6H_{12}O)_n

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

3 0
2 years ago
A closed container holds 2.0 moles of CO2 gas at STP. How many moles of oxygen can be placed in a container of the same size at
PilotLPTM [1.2K]

Answer: 2 moles

Explanation:

STP is Standard Temperature and Pressure. That means the pressure is 1.00 atm and the temperature is 273K. Since the oxygen is placed in the same container, we can use Ideal Gas Law to figure out what container the CO₂ used.

Ideal Gas Law: PV=nRT

P=1.00 atm

n=moles

R=0.08206 Latm/Kmol

T=273K

CO₂

V=\frac{nRT}{P}

V=\frac{(2.0 mol)(0.08206Latm/Kmol)(273K)}{1.00atm}

V=44.8L

Since we know that CO₂ has a 44.8 L container, we can use that to find the moles of oxygen.

n=\frac{PV}{RT}

n=\frac{(1.00atm)(44.8L)}{(0.08206Latm/Kmol)(273K)}

n=1.99=2mol

There are 2 mol of oxygen.

5 0
3 years ago
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