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3 years ago

8

When energy is transformed, some is wasted. What is a common form of wasted energy in some energy transformations? (Think about

inefficient light bulbs or car engines)

1 answer:

tresset_1 [31]3 years ago

4 0

Answer:

Eletrical

Explanation:

The electric transfers 70% of the input energy to kinetic energy 30% is wasted output energy in the form of thermal energy and sound.

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According to the collision theory of chemical reactions, an increase in the number of effective reactant collisions per unit tim

olasank [31]

1. true, more effective collisions per second. Faster reaction
2. <span>How can the reaction be slowed down? (talking about how fast or slow the reaction is)
</span>3. True
4. Rate lower if surface area decrease
5. Fine powder form
6. True
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Image questions:
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2. Energy of reactants higher than products.

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3 years ago

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How many moles of nitrogen gas will occupy a volume of .5L at 1.0atm and 298K

hammer [34]

<span>Use the Ideal law Equation :

P.V= n.R.T

V = 0.5 L

P = 1.0 atm

</span><span>R= 0.0821 L*atm/mol*K
</span>
<span>n = R*T/P*V

</span><span>P*V= n*R*T
</span>
1.0 * 0.5 = n *<span>0.0821*298

0,5 = n* 24.4658

n = 0,5 / 24.4658

n =0.0204 moles

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8 0

3 years ago

How should the enthalpy of an intermediate step be manipulated when used to produce an overall chemical equation?

Lapatulllka [165]

<span>The enthalpy of an intermediate step when used to produce an overall chemical equation should be manipulated in this way:
</span><span>Multiply the enthalpy by –1 if the chemical equation is reversed.
If the forward reaction requires energy, the reverse will produce energy.</span>

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3 years ago

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How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add

Nostrana [21]

Answer :

The correct answer for Mass of Na₂HPO₄ = 4.457 g and mass of NaH₂PO₄ = 8.23 g

Given : pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

<u>Step 1 : To find pka </u>

H₂PO₄⁻ <=> HPO₄²⁻

The above reaction has pka = 7.2 ( from image shown )

<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.

Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :

$pH = pka + log\frac{[A^-]}{[HA]}$

pH = 6.86 pKa = 7.2

$6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Subtracting both side by 7.2

$6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$-0.34 = log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Removing log

$10^-^0^.^3^4 = \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$ ---------------- equation (1)

<u>Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻</u>

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence, [H₂PO₄⁻ ] + [ HPO₄²⁻ ] = 0.1 M

Assume [H₂PO₄⁻ ] = x

So , [x ] + [ HPO₄²⁻ ] = 0.1 M

[ HPO₄²⁻ ] = 0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ]

[H₂PO₄⁻ ] = x

[ HPO₄²⁻ ] = 0.1 - x

Equation (1) = > $\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$

Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :

$\frac{[0.1 - x ]}{[ x]} = 0.457$

Cross multiplying

$0.1 - x = 0.457 x$

Adding x on both side

$0.1 -x + x = 0.457 x + x$

$0.1 = 1.457 x$

Dividing both side by 1.457

$\frac{0.1}{1.457} = \frac{1.457 x }{1.457}$

x = 0.0686 M

Hence , [H₂PO₄⁻ ] = x = 0.0686 M

[ HPO₄²⁻ ] = 0.1 - x

[ HPO₄²⁻ ] = 0.1 - 0.0686

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M or 0.0314 mole per 1 L

Since that volume of buffer solution is 1 L , so Molarity = mole

Hence Mole of NaH₂PO₄ = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

<u>Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄ </u>

Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :

Molar mass of Na₂HPO₄ = $141.96 \frac{g}{mol}$

Molar mass of NaH₂PO₄ = $119.98 \frac{g}{mol}$

$Mass (g) = mole (mol)* molar mass(\frac{g}{mol})$

$Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}$

Mass of Na₂HPO₄ = 4.457 g

$Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}$

Mass of NaH₂PO₄ = 8.23 g

5 0

3 years ago

Change these word equations into formula equations and balance them. Be sure to use the proper symbols to indicate the state of

Ket [755]

Answer:

See detailed reaction equations below

Explanation:

a) Mg(s) +2HBr(aq) ----------------> MgBr2(aq) + H2(g)

b) Ca(ClO3)2(s) ------------> CaCl2(s) + 3O2(g)

c) 3BaBr2(s) +2Na3PO4(aq) ------------> Ba3(PO4)2(s) + 6NaBr(aq)

d) 3AgNO3(aq) + AlI3(aq) --------------> 3AgI(s) + Al(NO3)3(aq)

Balancing reaction equations involves taking valencies and number of atoms of each element on the reactants and products side into consideration respectively.

3 0

3 years ago