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Sedbober [7]
3 years ago
7

Static charges can be applied to neutral objects by friction, induction or conduction. What do all of these methods utilize to c

reate this charge?
a.protons
b.electrons
c.heat energy
Chemistry
1 answer:
MatroZZZ [7]3 years ago
5 0
Electrons are valence and free moving so they take place in charge transfer
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The energy released by nuclear fission can be used to ________ and turn a turbine to produce ________ energy.
enyata [817]
Heat water; mechanical (the movement of a turbine is based off of mechanical energy, not chemical or potential).
6 0
3 years ago
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What is the temperature of 0.460 mol of gas at a pressure of 1.2 atm and a volume of 12.5 L ? Express your answer using two sign
Taya2010 [7]

The temperature is 370K.

The volume of a given fuel pattern is immediately proportional to its absolute temperature at regular pressure (Charles's law). The volume of a given amount of fuel is inversely proportional to its pressure whilst temperature is held steady (Boyle's regulation).

Density is immediately proportional to stress and indirectly proportional to temperature. As stress increases, with temperature constant, density will increase. Conversely when temperature increases, with strain regular, density decreases.

The equations describing those legal guidelines are unique cases of the best gasoline regulation, PV = NRT, wherein P is the pressure of the gas, V is its extent, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (common) gas constant.

Learn more about pressure here: brainly.com/question/25736513

#SPJ4

5 0
2 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
How many moles of H2O are needed to produce 5.6 mol of NaOH?<br> Na2O + H2O --&gt; 2NaOH
navik [9.2K]

Answer: 2.8 moles

Explanation:

The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.

Na2O + H2O --> 2NaOH

1 mole of H2O = 2 moles of NaOH

Let Z moles of H2O = 5.6 mole of NaOH

To get the value of Z, cross multiply

5.6 moles x 1 mole= Z x 2 moles

5.6 = 2Z

Divide both sides by 2

5.6/2 = 2Z/2

2.8 = Z

Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH

3 0
3 years ago
A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *
elena-14-01-66 [18.8K]

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

  1. P1 = initial pressure (kPa)
  2. P2 = final pressure (kPa)
  3. V1 = initial volume (L)
  4. V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

3 0
3 years ago
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