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vesna_86 [32]
3 years ago
10

Electronic and electrical devices differ in the way they use _____.

Chemistry
1 answer:
Katena32 [7]3 years ago
3 0
Election current because voltage is a measurement, information doesn't apply to all electrical devices and the wires within are usually copper bc it conducts and hardly ever will the wires be anything different because copper is cheap
You might be interested in
How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
3 years ago
Write the formulas of the following compounds:
dem82 [27]

Answer:

a) Li2CO3

b) NaCLO4

c) Ba(OH)2

d) (NH4)2CO3

e) H2SO4

f) Ca(CH3COO)2

g) Mg3(PO4)2

f) Na2SO3

Explanation:

a) 2Li + CO3 ↔ Li2CO3

b) NaOH * HCLO4 ↔ NaCLO4 + H2O

c) Ba + 2H2O ↔ Ba(OH)2 +

d)  2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O

c) SO2 + NO2 +H2O ↔ H2SO4 + NOx

f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O

g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O

h) NaOH + H2SO3 ↔ Na2SO3 + H2O

6 0
3 years ago
What is the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form sol
rusak2 [61]

Answer:

PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)

Explanation:

Hello!

In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)

Best regards!

7 0
3 years ago
7. How many moles of mercury(ii) oxide, hgo, are needed to produce 12. 5 g of oxygen, o2? 2 hgo(s) ---&gt;2 hg(l) o2(g)
svetoff [14.1K]

Moles are the division of the mass and the molar mass. The moles of mercury (ii) oxide in the decomposition reaction needed to produce oxygen are 0.781 moles.

<h3>What is a decomposition reaction?</h3>

A decomposition reaction is a breakdown of the reactant into simpler products. The decomposition of mercury (ii) oxide can be shown as:

2HgO(s) → 2Hg(l) + O₂(g)

From the reaction, it can be said that 2 moles of mercury (ii) oxide decomposes to produce 1 mole of oxygen.

The moles of oxygen that needs to be produced are calculated as:

Moles = mass ÷ molar mass

= 12.5 gm ÷ 32 gm/mol

= 0.39 moles

0.39 moles of oxygen are needed to be produced.

From the stoichiometric coefficient of the reaction, the moles of HgO is calculated as: 2 × 0.39 = 0.781 moles

Therefore, 0.781 moles of HgO are required in the reaction.

Learn more about moles here:

brainly.com/question/3801333

#SPJ4

5 0
2 years ago
Need help ASAP!<br><br> How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?
Setler79 [48]

Answer:

\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

  • molarity= 1.2 mol NaNO₃/L
  • liters of solution=0.25 L
  • moles of solute =x

1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L

0.25 \ L *1.2 \ mol \ NaNO_3/L=x

The units of liters cancel, so we are left with the units moles of sodium nitrate.

0.25  *1.2 \ mol \ NaNO_3=x

0.3 \ mol \ NaNO_3=x

There are 0.3 moles of sodium nitrate.

3 0
2 years ago
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