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3 years ago

When Philip analyzed unknown substance x, he found that it contained only one kind of atom what is substance x

1 answer:

5 0

Substance X is a (B) Element.
Explanation:

A compound consists of either many identical atoms or more than 1 type of atom. They can be broken down or converted into a simpler one.
An element is a substance which contains only one kind of atom. It cannot be broken down further or converted into simpler substance by any chemical means.

Heterogeneous mixture is a non uniform mixture of different atoms without any chemical bonding in between them.
Homogeneous mixture is a uniform mixture of different atoms without any chemical bonding in between them.
From the definitions it is clear that substance X is (B) Element.

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Write the formula for calcium chloride. Explain how you determined this formula

hichkok12 [17]

Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is <span>CaC<span>l2</span></span>

Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions.

To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.

7 0

3 years ago

A fluorine atom has 9 positive charge,9 negative charges and a mass of 19. Describe the structure of its atom.

Tanya [424]

The atomic structure of the atom contains 9 positively charged particles (protons) and 10 neutrally charged particles (neutrons) in the center of the atom in a clump called the nucleus. Those 9 negatively charged particles (electrons) are moving around outside of the nucleus.
There are 10 neutral charges, because the mass of 19 comes from the number of neutral charges plus the number of positive charges.
To calculate the number of neutral charges, subtract the positive charges from the mass (19 - 9), and you get the number of neutral charges (10).

6 0

3 years ago

Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag. Of the two metals, copper

KatRina [158]

Answer:

Answer is given below.

Explanation:

Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.

In cell representation, half cell present left to salt-bridge notation $(\parallel )$ is anodic system and another half cell present right to salt-bridge notation $(\parallel )$ is cathodic system.

So anode is Cu and cathode is Ag.

oxidation: $Cu-2e^{-}\rightarrow Cu^{2+}(aq.)$

[reduction: $Ag^{+}+e^{-}\rightarrow Ag$ ] $\times 2$

-----------------------------------------------------------------------------------------------

chemical equation: $Cu+2Ag^{+}(aq.)\rightarrow Cu^{2+}(aq.)+2Ag$

Oxidizing agent is that species which takes electron from another species. Here $Ag^{+}(aq.)$ takes electron from Cu. Hence $Ag^{+}(aq.)$ is the oxidizing agent.

Reducing agent is that species which gives electron to another species. Here Cu gives electron to $Ag^{+}(aq.)$ . Hence Cu is the reducing agent.

6 0

3 years ago

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('

polet [3.4K]

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$

Next, by using the density of platinum we compute the volume:

$V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3$

Which computed in terms of the edge length is:

$V=a^3$

Therefore, the edge length turns out:

$a=\sqrt[3]{180cm^3}\\ \\a=5.65cm$

Best regards.

6 0

3 years ago

A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First

bekas [8.4K]

Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

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

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

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

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0

3 years ago