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Vinvika [58]
3 years ago
12

Zinc is more active than cobalt and iron but less active than aluminum. Cobalt is more active than nickel but less active than i

ron. Which of the following correctly lists the elements in order of increasing activity?
a) Zn < Al < Co < Ni < Fe
b) Co < Ni < Fe < Zn < Al
c) Fe < Ni < Co < Al < Zn
d) Ni < Co < Fe < Zn < Al
e) Ni < Fe < Co < Zn < Al
Chemistry
1 answer:
mixas84 [53]3 years ago
5 0

Answer:

Option D

Explanation:

For the first statement, the order is:

=> Co < Fe < Zn < Al

For the second statement it is,

=> Ni < Co < Fe

So, combining the order becomes:

=> Ni < Co < Fe < Zn < Al

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Copper produces a green flame test. When is the green light emitted?​
belka [17]

Answer: When you burned the skewer tip coated with copper sulfate the green light is emitted

Explanation:

when the copper is burned it makes a green light

8 0
3 years ago
What is the % composition of Carbon in Chromium (iii) Carbonate
photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

\begin{gathered} C\rightarrow3\times12=36 \\  \\ O\rightarrow9\times16=144 \\  \\ Cr\rightarrow2\times52=104 \end{gathered}

The molar mass will be thus:

M=36+104+144=284\text{ g/mol}

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\  \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}

The percent composition of Carbon is thus 12.7 %.

8 0
1 year ago
Maggie is a member of her school’s environmental club and is interested in recycling. She asks the question, “How does exposure
miv72 [106K]
The best describes on why and how Maggie should change her question to make it a better scientific question is that t<span>he question involves giving an opinion about recycling, so it should be changed to rely only on facts. Hope it helps! </span>
8 0
3 years ago
Read 2 more answers
I need the answer today!
olchik [2.2K]

_____________________________________________

<h2 /><h2 /><h2><u>Solution </u><u>3</u><u> Is The Most Concentrated</u></h2>

<h3>S1:</h3>

M = m/v

= 100ml ÷ 2 spoons × 100%

= - 5,000 μg/ppb³

= <u>50% Diluted</u>

<h3>S2:</h3>

M = m/v

= 200ml ÷ 5 spoons × 100%

= - 4,000 μg/ppb³

= <u>40% Diluted</u>

<h3>S3:</h3>

M = m/v

= 300ml ÷ 6 spoons × 100%

= - 5,000 μg/ppb³

= <u>50% Diluted</u>

<h3>S4: </h3>

M = m/v

= 600ml ÷ 8 spoons × 100%

= - 75,000 μg/ppb³

= <u>75% Diluted</u>

_____________________________________________

8 0
3 years ago
Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressure
Volgvan

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of O_2 and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = n_1=\frac{51.8}{32 g/mol}=1.619 mol

Moles of helium gas = n_2=\frac{33.1 g}{4 g/mol}=8.275 mol

Total moles of gas = n_1+n_2=(1.619 +8.275 ) mole=9.894 mol

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

P=\frac{nRT}{V}

=\frac{9.894 mol\times 0.0821 atm L/mol K\times 294 K}{10.0 L}

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

p_{1}=P\times \chi_{1}=P\times \frac{n_1}{n_1+n_2}

p_1=23.88 atm\times \frac{1.619 mol}{9.894 mol}=3.91 atm

Partial pressure of the helium gas

p_{2}=P\times \chi_{2}=P\times \frac{n_2}{n_1+n_2}

p_2=23.88 atm\times \frac{8.275 mol}{9.894 mol}=19.97 atm\approx 20.0 atm

6 0
3 years ago
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