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Galina-37 [17]
3 years ago
9

Good science does not depend on interactions within the scientific community. True False

Chemistry
2 answers:
Shkiper50 [21]3 years ago
7 0

The answer would be true!

I hope this helps!

Aloiza [94]3 years ago
4 0
I believe the answer is true
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How many grams of chromium are in 2.0 x 1024 atoms
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Answer: 173 g ( 0.17 kg in right accuracy)

Explanation: Amount in moles is n = N/Na =  2.0·10^24 / 6.022·10^23 (1/mol).

n = 3.32116 mol. M(Cr) = 52.00 g/mol and  mass m = nM = 172.7 g

7 0
3 years ago
at stp which of following would have the same number of molecules a 1 l of c2h4 gas? a. 0.5 of H2 b. 1L of Ne c.2L of H2O d.3L o
sashaice [31]

Answer:d

Explanation:

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DNA is a nucleic acid made of building blocks called nucleotides.
emmasim [6.3K]

Answer:

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2 years ago
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How many moles of H,O must be decomposed to form 200 moles of H,?<br> 24,0 2H +0,
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8 0
3 years ago
The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
Karolina [17]

<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 3.45

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

8 0
3 years ago
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