Can some balance the equation? FeBr3 + H2SO --> Fe2(SO4)3 + HBr

A weather balloon is filled with helium that occupies a volume of 5.00 × 10^4 L at 0.995 atm and 32.0°C. After it is released, i

DaniilM [7]

Answer:

The new volume is 5.913*10^4 L

Explanation:

Step 1: Write out the formula to be used:

Using general gas equation;

P1V1 / T1 =P2V2 /T2

V2 = P1V1T2 / P2T1

Step 2: write out the values given and convert to standard unit's where necessary

P1 = 0.995atm

P2 0.720atm

V1 = 5*10^4 L

T1 = 32°C = 32+ 273 = 305K

T2 = -12°C = -12 + 273 = 261K

Step 3: Equate your values and do the calculation:

V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305

V2 = 1298.475 * 10^4 / 219.6

V2 = 5.913 * 10^4 L

So the new volume of the balloon is 5.913*10^4 L

3 0

3 years ago

Match the sentences with the steps of the scientific method.

Harrizon [31]

Answer: use your brain

Explanation:

I have one

6 0

2 years ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0

2 years ago

Which metals may be oxidized by H+ under standard-state conditions? Ag+(aq) + e– → Ag(s) E° = 0.80 V Cu2+(aq) + 2e– → Cu(s) E° =

Debora [2.8K]

Answer is: tin and zinc, because they standard potential as less than zero.
Tin and zinc are oxidized to tin and zinc cations (with +2 charge) and hydrogen anions are reduced to hydrogen molecules with neutral charge.
Zn → Zn²⁺ + 2e⁻; 2H⁺ + 2e⁻ → H₂.
Oxidation is increase of oxidation number and reduction is decrease of oxidation number.

8 0

2 years ago

I just need a sentence

irina [24]

Answer:

Plankton will never get the kraby patty secret formula

Explanation:

7 0

3 years ago

Read 2 more answers

Can some balance the equation?FeBr3 + H2SO --> Fe2(SO4)3 + HBr​

Can some balance the equation?FeBr3 + H2SO --> Fe2(SO4)3 + HBr