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3 years ago

According to the dichotomous key, organism 2 is a

Chemistry

2 answers:

Ulleksa [173]3 years ago

6 0

Answer:

water scorpion

Explanation:

good luck have a nice day

Bogdan [553]3 years ago

3 0

I think its watee scorpion

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Which of these expressions are correct variations of the Combined Gas Law?

mafiozo [28]

Answer:

Both

Explanation:

The combined gas law is also known as the general gas law.

From the ideal gas law we assume that n = 1;

So;

PV = nRT

and then;

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

If we cross multiply;

P₁V₁T₂ = P₂V₂T₁

So;

T₁ = $T_{2} \frac{P_{1}V_{1} }{P_{2} V_{2} }$

Also;

V₂ = $V_{1} \frac{P_{1} T_{2} }{P_{2} T_{1} }$

So from the choices both are correct

3 0

3 years ago

Why is salt water intrusion a big problem in florida?

Schach [20]

Its leading to contaminated drinking water.

7 0

3 years ago

Read 2 more answers

Which is NOT a chemical reaction?

Sophie [7]

A is correct because there is no reaction involved

6 0

3 years ago

Read 2 more answers

A chemical reaction produced 10.1 cm3 of nitrogen gas at 23 °C and 746 mmHg. What is the volume of this gas if the temperature a

adell [148]

Answer:

volume of gas = 9.1436cm³

Explanation:

We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.

Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.

V1 =10.1cm³ , P1 =746mmHg, T1=23°C =23+273=296k

V2 =? , P2 =760mmmHg , T2=0°C = 0+273 =273K

Using the combined gas equation to calculate for V2;

$\frac{V1P1}{T1}=\frac{V2P2}{T2} \\ re-arranging, \\V2 =\frac{V1P1T2}{P2T1}$

$V2 =\frac{10.1*746*273}{760*296}$

V2=9.1436cm³

6 0

4 years ago

The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0

3 years ago