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3 years ago

Which substance particle will be attracted by a positively charged object

Chemistry

1 answer:

laiz [17]3 years ago

5 0

Answer:

Protons

Protons are another type of subatomic particle found in atoms. They have a positive charge so they are attracted to negative objects and repelled from positive objects.

Explanation:

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Which is an example of pure chemistry? a. shape of a soybean plant b. chemicals containing carbon c. finding an antidote

Rina8888 [55]

I believe that it is B chemicals containing carbon

5 0

3 years ago

How can there be more than 1000 different atoms when there are only about 100 different elements?

Irina-Kira [14]

Um im pretty sure there are only about 100 different atoms...

7 0

3 years ago

9. How many grams of potassium sulfate are needed to make 250 mL of a 0.150 M

antiseptic1488 [7]

Answer:

6.53g of K₂SO₄

Explanation:

Formula of the compound is K₂SO₄

Given parameters:

Volume of K₂SO₄ = 250mL = 250 x 10⁻³L

= 0.25L

Concentration of K₂SO₄ = 0.15M or 0. 15mol/L

Unknown:

Mass of K₂SO₄ =?

Methods:

We use the mole concept to solve this kind of problem.

>>First, we find the number of moles using the expression below:

Number of moles= concentration x volume

Solving for number of moles:

Number of moles = 0.25 x 01.5

= 0.0375mole

>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:

Mass(g) = number of moles x molar mass

Solving:

To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.

For:

K = 39g

S = 32g

O = 16g

Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)

= 78 +32 + 64

= 174g/mol

Using the expression:

Mass(g) = number of moles x molar mass

Mass of K₂SO₄ = 0.0375 x 174 = 6.53g

5 0

3 years ago

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

3 0

4 years ago

Which element has the electron configuration [Xe] 6s2 4f14 5d10 6p2?

andreev551 [17]

To determine the element that has an electron configuration of <span>[Xe] 6s2 4f14 5d10 6p2, we must know the atomic number of Xe. From the periodic table, Xe has an atomic number of 54. We add the proceeding electrons, which is now a total of 82. The element with this atomic number is lead, Pb.</span>

4 0

3 years ago