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Tanzania [10]
4 years ago
11

Hey um i dont get this yall wont to help me out???

Chemistry
1 answer:
den301095 [7]4 years ago
4 0
3. 9
20. 60 (1hour)
To get to 60 from 20 we multiply by three. 3 times 3 is 9.
A is true
B is true
C is true
D is true
E is false
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which statement besy describes how the calormeter can be used to determine ghe specific heat capacity of the metal sample
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Energy transfers to the metal from the water and calorimeter until they are all at room temperature.

Explanation:

i hope this helps

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4 years ago
What 3 gases does the flame test check for and what should you expect to see/hear?
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What of the gasses is N and the flame changes it's color from Orange to blue
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3 years ago
Calculate the enthalpy of the formation of butane, C4H10, using the balanced chemical equation and the standard value below:
zavuch27 [327]

Answer:

+125.4 KJmol-1

Explanation:

∆H C4H10(g) = -2877.6kJ/mol

∆H C(s)=-393.5kJ/mol

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∆H reaction= ∆Hproducts - ∆H reactants

∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]

∆H reaction= +125.4 KJmol-1

6 0
3 years ago
Unrestricted populations of organisms experience _____.
Fudgin [204]
Unrestricted populations of organisms experience Exponential growth. 

As a population reaches is carrying capacity there is an increase in competition for All of the above.
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5 0
3 years ago
Read 2 more answers
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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