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2 years ago

Please help me for real

Chemistry

1 answer:

Bond [772]2 years ago

7 0

Answer:

I believe it is the replication fork. so B

Explanation:

I haven't done this section in quite some time. this was biology for me, and I'm in chem right now. so I hope that's right. I hope I could help you :)

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Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass

Ipatiy [6.2K]

Answer:

1.8g

Explanation:

Initial volume = 43.5ml

Final volume = 49.4ml

Mass = 10.88g

Density = ?

Volume = Final volume - initial volume

= 49.4 - 43.5

= 5.9ml

Density = Mass/volume

Density = 10.88/5.9

= 1.8g/ml

4 0

3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0

2 years ago

In a chemical reaction, ____ interact to form ____. select one:

QveST [7]

A. Reactants, products

In a chemical reaction, reactants interact to form products

7 0

3 years ago

Net ionic equation for Calcium hydroxide reacting with ammonium Chloride?

VMariaS [17]

Answer:

Ca(OH)₂ (aq) + 2NH₄Cl (aq) → CaCl₂ (aq) + 2NH₃ (g) + 2H₂O (l)

Ca²⁺ + 2OH⁻ + 2NH₄⁺ + 2Cl⁻ → Ca⁺² + 2Cl⁻ + 2NH₃ (g) + 2H₂O (l)

2OH⁻ + 2NH₄⁺ → 2NH₃ (g) + 2H₂O (l)

3 0

1 year ago

Read 2 more answers

a heliox tank contains 32% helium and 68% oxygen. the total pressure in the tank is 395 kPa. What is the partial pressure of oxy

trapecia [35]

Answer:

Pp O2 = 82.944 KPa

Explanation:

heliox tank:

∴ %wt He = 32%

∴ %wt O2 = 68%

∴ Pt = 395 KPa

⇒ Pp O2 = ?

assuming a mix of ideal gases at the temperature and volumen of the mix:

∴ Pi = RTni/V

∴ Pt = RTnt/V

⇒ Pi/Pt = ni/nt = Xi

⇒ Pi = (Xi)*(Pt)

∴ Xi: molar fraction (ni/nt)

⇒ 0.68 = mass O2/mass mix

assuming mass mix = 100 g

⇒ mass O2 = 68 g

∴ molar mass O2 = 32 g/mol

⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2

⇒ mass He = 32 g

∴ molar mass He = 4.0026 g/mol

⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He

⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles

molar fraction O2:

⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099

⇒ Pp O2 = (X O2)(Pt)

⇒ Pp O2 = (0.2099)(395 KPa)

⇒ Pp O2 = 82.944 KPa

6 0

3 years ago