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Vedmedyk [2.9K]
3 years ago
5

What are the types of Asexual reproduction? need this asap bro uyftghjkddfs

Chemistry
1 answer:
kolezko [41]3 years ago
5 0

Answer:

The types of Asexual reproduction include:  fission, fragmentation, budding, vegetative reproduction, spore formation and agamogenesis.

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A flask contains 0.83mol of neon gas at a temperature of 35°C. The pressure gauge indicates 0.37atm inside the flask. What is th
Angelina_Jolie [31]

Answer: 5.66 dm3

Explanation:

Given that:

Volume of neon gas = ?

Temperature T = 35°C

Convert Celsius to Kelvin

(35°C + 273 = 308K)

Pressure P = 0.37 atm

Number of moles N = 0.83 moles

Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

0.37atm x V = 0.83 moles x 0.0082 atm dm3 K-1 mol-1 x 308K

0.37 atm x V = 2.096 atm dm3

V = (2.096 atm dm3 / 0.37atm)

V = 5.66 dm3

Thus, the volume of the neon gas is 5.66 dm3

5 0
3 years ago
Can you help me please
den301095 [7]
Type in the Eye lins it will help
4 0
3 years ago
Agree or disagree? Explain (2 ideas, 2 examples)
yawa3891 [41]

Answer:

Agree this is correct if it not blame me

6 0
3 years ago
How many equivalents are present in 10 g of Ca2+? <br> 1 <br> 0.5 <br> 1.5 <br> 2
Lynna [10]

0.5

Explanation:

Given parameters:

Mass of Ca²⁺ = 10g

unknown:

Equivalent weight = ?

Solution:

Equivalent weight that is the amount of electrons which a substance gains or loses per mole.

Ca²⁺ has +3 charge

It lost 2e⁻;

therefore;

  In 1 mole of  Ca²⁺, we have 2 equivalent weight

1 mol  Ca²⁺ = 2eq. wts. 

1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt. 

Therefore;

10.0 g  Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.

learn more:

Molar mass brainly.com/question/2861244

#learnwithBrainly

8 0
3 years ago
Cyclopropene decomposes to propene when heated to 500 C, calculate rate constant for the first order reaction 0 min 1.48 mmol/L
Dmitry_Shevchenko [17]

Answer:

k = 0.0306 min-1

Explanation:

The table is given as;

Time, Concentration

0 1.48

5 1.27

10 0.98

15 0.84

The integrated rate law for a first order reaction is given as;

ln [A] = -kt + ln [Ao]

where;

[A] = Final Concentration

[Ao] = Initial Concentration

k = rate constant

t = time

In the table, taking the first two sets of values;

t = 5

k = ?

[Ao]  = 1.48

[A] = 1.27

Inserting into the equation;

ln(1.27) = - k (5) + ln(1.48)

ln(1.27)  - ln(1.48) = -5k

-0.1530 = -5k

k = -0.1530 / -5

k = 0.0306 min-1

6 0
3 years ago
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