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matrenka [14]
3 years ago
9

What hold's the nucleus together in an atom?

Chemistry
1 answer:
Simora [160]3 years ago
6 0
The nuclear force holds the nucleus together in an atom
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Can someone help me out with this?
ycow [4]

Answer:

2.8 * 10^(-6) / 1.4 * 10^(-2)=

2* 10^(-8)

4 0
4 years ago
I need help with this anyone?
valkas [14]
Just look it up on goog^le or a chart
5 0
3 years ago
Q2. Which formula represents a homogeneous mixture?
bija089 [108]

Answer:

D) HCl(aq)

Explanation:

A homogeneous mixture can be defined as any liquid, solid or gaseous mixture which has an identical or uniform composition and properties throughout any given sample of the mixture. In Chemistry, all solutions are considered to be a homogeneous mixture.

In this scenario, the chemical formula which represents a homogeneous mixture is aqueous hydrogen chloride, HCl(aq). The aqueous hydrogen chloride is a homogeneous mixture of water and hydrogen chloride. This ultimately implies that, aqueous hydrogen chloride HCl(aq) is a solution of hydrogen chloride in water and it is commonly referred to as Hydrochloric acid.

Given by the chemical equation;

HCl(aq) + H_{2}O(l)  →  H_{3}O^{+} (aq) + Cl_{−}(aq)

8 0
3 years ago
I need help, plz help me with this problem
Svetlanka [38]

Answer:

It's b

Explanation:

I had the same exact question

5 0
2 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
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