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Mashcka [7]
3 years ago
5

How do combination circuits help prevent problems in circuits in a home? If it helps, think about a bedroom since all electrical

work in a bedroom is on the same circuit.
Chemistry
1 answer:
den301095 [7]3 years ago
6 0
‍♀️ ◻️✅◾️▫️‍◾️◻️◻️◼️
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In multicellular organismscells that become specialized to perform a specific function are called?
Zolol [24]
The cell proliferates to produce many cells that result in multicellar organism.

answer
8 0
2 years ago
PLEASE HELP I NEG YOU FATHERCARRAS PLEASE HELP ME !!!
LekaFEV [45]
Answer 
<span>How does adding a non-volatile solute to a pure solvent affect the vapor pressure of the pure solvent?

</span>Answer The third option The solvent's vapor pressure will not be affected.

<span>To make a 2.0 M solution, how many moles of solute must be dissolved in 0.50 liters of solution?

Answer C. 1.0 mole solute.
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3 0
3 years ago
Why isnt my grandma talking to me
Rufina [12.5K]

Answer:

She saw you at the elderly N*de swimming event.

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4 0
2 years ago
Combustion of ethylene glycol leads to a change in internal energy (\DeltaU) of –1189 kJ at 298 K. What is the corresponding cha
ddd [48]

Answer:

-1190.24 kJ

Explanation:

The enthalpy change in a chemical reaction that produces or consumes  gases is given by the expression:

ΔH = ΔU + Δngas RT

where  Δn gas is the change of moles of gas, R is the gas constant,and T is temperature.

Now from the given   balanced chemical reaction, the change in number of mol gas is equal to:

Δn gas = mole gas products - mole gas reactants =  2 - 5/2 = -1/2 mol

Sionce we know ΔU and the temperature (298 K), we are in position to calculate the change in enthalpy.

ΔH = -1189 x 10³ J + (-0.5 mol ) 8.314 J/Kmol x 298 K

ΔH = -1.190 x 10⁶ J = -1.190 x 10⁶ J x 1 kJ/1000 J = -1.190 x 10³ J

7 0
3 years ago
If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L
Tom [10]

Answer:

Approximately 2.53\; \rm L (rounded to three significant figures) assuming that {\rm HCl}\, (aq) is in excess.

Explanation:

When {\rm HCl} \, (aq) and {\rm AgNO_3}\, (aq) precipitate, {\rm AgCl} \, (s) (the said precipitate) and \rm HNO_3\, (aq) are produced:

{\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq) (verify that this equation is indeed balanced.)

Look up the relative atomic mass of \rm Ag and \rm Cl on a modern periodic table:

  • \rm Ag: 107.868.
  • \rm Cl: 35.45.

Calculate the formula mass of the precipitate, \rm AgCl:

\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm  g\cdot mol^{-1}\end{aligned}.

Calculate the number of moles of \rm AgCl formula units in 45.00\; \rm g of this compound:

\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}.

Notice that in the balanced equation for this reaction, the coefficients of {\rm AgNO_3} \, (aq) and {\rm AgCl}\, (s) are both one.

In other words, if {\rm HCl}\, (aq) (the other reactant) is in excess, it would take exactly 1\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 1\; \rm mol \! of {\rm AgCl}\, (s)\! formula units.

Hence, it would take 0.313987\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 0.313987\; \rm mol\! of {\rm AgCl}\, (s)\! formula units.

Calculate the volume of the {\rm AgNO_3} \, (aq)\! solution given that the concentration of the solution is 0.124\; \rm mol \cdot L^{-1}:

\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}.

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of {\rm AgNO_3} \, (aq)\!.)

In other words, approximately 2.53\; \rm L of that {\rm AgNO_3} \, (aq)\! solution would be required.

4 0
3 years ago
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