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2 years ago

Which of the following is a single replacement reaction?

Chemistry

2 answers:

Morgarella [4.7K]2 years ago

4 0

Fe + 3NaBr → FeBr3 + 3Na

The Na is replaced by the Fe atom.

MArishka [77]2 years ago

4 0

The correct answer is D) Fe + 3NaBr → FeBr3 + 3Na

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Something that can be broken down by living organisms is

Aloiza [94]

The answer is A: biodegradable.

5 0

2 years ago

Mass

DiKsa [7]

The molecular formula =C₆H₁₂O₆

<h3>Further explanation</h3>

Given

6.00 g of a certain compound X

The molecular molar mass of 180. g/mol

CO₂=8.8 g

H₂O=3.6 g

Required

The molecular formula

Solution

mass C in CO₂ :

= 1.12/44 x 8.8

= 2.4 g

mass H in H₂O :

= 2.1/18 x 3.6

= 0.4 g

Mass O in compound :

= 6-(2.4+0.4)

= 3.2 g

Mol ratio C : H : O

= 2.4/12 : 0.4/1 : 3.2/16

= 0.2 : 0.4 : 0.2

= 1 : 2 : 1

The empirical formula : CH₂O

(CH₂O)n=180 g/mol

(12+2+16)n=180

(30)n=180

n=6

(CH₂O)₆=C₆H₁₂O₆

7 0

2 years ago

State the location and the total charge of the protons in a Ni-64 atom.

blsea [12.9K]

Location = nucleus ,
and you can see the total charge on the number that written below the substance, which would be 28+

hope this helps

6 0

3 years ago

Read 2 more answers

The mass percent of carbon in a typical human is 18%, and the mass percent of 14C in natural carbon is 1.6 × 10-10%. Assuming a

Tema [17]

Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

$\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g$

Carbon-14 is $1.6\times 10^{-10}\ \%$ of the carbon in the body. So,

$\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g$

Also,

14 g of Carbon-14 contains $6.023\times 10^{23}$ atoms of carbon-14

So,

$2.48\times 10^{-8}\ g$ of Carbon-14 contains $\frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8}$ atoms of carbon-14

Atoms of carbon-14 = $1.07\times 10^{15}$

Given that:

Half life = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.00012 years⁻¹

Also, 1 year = $3.154\times 10^7$ s

So, The rate constant, k = $\frac{0.00012}{3.154\times 10^7}$ s⁻¹ = $3.8\times 10^{-12}\ s^{-1}$

Thus, decay events per second = $K\times atoms decayed$ = $3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s$ = 4066 decay/s

4 0

3 years ago

¿¿¿Alguien sabe que sustancias no concuerdan con el modelo de Arrhenius??? Es urgente, ayuden por favor

Kay [80]

Twjegnefnwfnnegengefbefb no se

8 0

3 years ago