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Law Incorporation [45]
3 years ago
14

Two unknown elements react to form a single compound. The compound is shiny and malleable. What would you expect to be true abou

t the compound produced?
A.
The compound will probably have properties of both elements like a metalloid.

B.
The compound will probably have high boiling points and melting points and will be conductive in solution.

C.
The compound will probably have low boiling and melting points and will not be conductive in solution.

D.
The compound will probably have high boiling and melting points and will not be conductive in solution.
Chemistry
2 answers:
Fed [463]3 years ago
7 0
I believe the answer is B
mamaluj [8]3 years ago
3 0

Answer:

B

Explanation:

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HELP.
Kamila [148]

Answer:

The mass of tin is 164 grams

Explanation:

Step 1: Data given

Specific heat heat of tin = 0.222 J/g°C

The initial temeprature of tin = 80.0 °C

Mass of water = 100.0 grams

The specific heat of water = 4.184 J/g°C

Initial temperature = 30.0 °C

The final temperature = 34.0 °C

Step 2: Calculate the mass of tin

Heat lost = heat gained

Qlost = -Qgained

Qtin = -Qwater

Q = m*c*ΔT

m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)

⇒with m(tin) = the mass of tin = TO BE DETERMINED

⇒with c(tin) = the specific heat of tin = 0.222J/g°C

⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C

⇒with m(water) = the mass of water = 100.0 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C

m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C

m(tin) =  163.9 grams ≈ 164 grams

The mass of tin is 164 grams

4 0
3 years ago
According to the law of conservation of matter, we know that the total number of atoms does not change in a chemical reaction an
Shkiper50 [21]

Answer:

Compounds are represented by chemical formulas.

Explanation:

3 0
2 years ago
Use this oxidation-reduction reaction to answer questions about half-reactions:
Advocard [28]

Answer:

N, O, R

Explanation:

Thanks for answering bro, you saved me

5 0
2 years ago
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

5 0
3 years ago
Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves
hjlf

This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)

Explanation :

The given two chemical reactions are:

(1) 2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

(2) 3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)

First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:

(1)

(2) 6MnO_2(s)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)

Now we are adding both the reactions, we get the overall chemical reaction.

6MnCO_3(s)+3O_2(g)+6MnO_2(s)+8Al(s)\rightarrow 6MnO_2(s)+6CO_2(g)+6Mn(s)+4Al_2O_3(s)

The  MnO_2 is common on both side, by cancelling it, we get:

The net chemical equation for the production of manganese is:

6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)

5 0
3 years ago
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