How would you solve this problem??

Answers

Answer 1

Answer: Calculate the pOH where [OH⁻]=4.7×10⁻³M:
pOH=-log[OH⁻]
pOH=-log(4.7×10⁻³)
pOH=2.33

Calculate [OH⁻] if the pOH is 1.34:
pOH=-log[OH⁻]
10^(-pOH)=[OH⁻]
[OH⁻]=10^(-pOH)
[OH⁻]=10^(-1.34)
[OH-]=0.0457M or [OH⁻]=4.57×10⁻²M

Calculate pH if the [OH⁻] is 1.74×10⁻²M:
[H⁺]=K(w)/[OH⁻] or pOH=-log[OH⁻]
[H⁺]=(1×10⁻¹⁴)/(1.57×10⁻²) pOH=-log(1.74×10⁻²)
[H⁺]=6.369×10⁻¹³M pOH=1.759
pH=-log[H⁺] pH=14-pOH
pH=-log(6.369×10⁻¹³) pH=14-1.759
pH=12.2 pH=12.2

I hope this helps. Let me know if anything is unclear.

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The sedimentary rock formed when water deposits tiny particles of clay in very thin flat layers is called?a.gymsum b.shale c.limestone d.calcite

OXIDATION IN
Cl2(aq) + 2Brmc003-1.jpg(aq) mc003-2.jpg 2Clmc003-3.jpg(aq) + Br2(aq)

Answers

I think the chemical reaction is:

Cl2(aq) + 2Br-(aq) = 2Cl-(aq) + Br2(aq)

The oxidation in this reaction is the reaction for bromine where the charge goes from -1 to a neutral charge while the reduction reaction is showed by the chlorine from a neutral charge to a -1 charge.

Which nuclear decay emission consists of energy, only?(1) alpha particle (3) gamma radiation
(2) beta particle (4) positron

Answers

Answer: (3) gamma radiation

Explanation:

An isotope can decay in 4 process:

1.) Alpha decay: In this process, alpha particles is emitted when a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.

$_Z^A\textrm{X}\rightarrow _(Z-2)^(A-4)+_2^4\alpha$

2.)Beta-decay: In this process, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

$_Z^A\textrm{X}\rightarrow _(Z+1)^A\textrm{Y}+_(-1)^0\beta$

3.) Gamma ray emission: in this process, an unstable nuclei gives off excess energy by a spontaneous electromagnetic process and releases $\gamma -radiations$ . These radiations does not carry any charge and are electrically neutral.

$_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma$

4.) Positron decay: In this process, a proton gets converted to neutron and an electron neutrino and releases positron particles. This particle carries a charge of +1 units.

$_Z^A\textrm{X}\rightarrow _(Z-1)^A\textrm{Y}+_(+1)^0e$

In the question, it is given that when an isotope decays, it emits only energy. It is released when an isotope undergoes gamma decay and therefore, the correct answer is gamma radiation.

(3) gamma radiationIt is Electromagnetic energy only

A package of aluminum foil is 63.2 yd long, 11 in. wide, and 0.00035 in. thick. If aluminum has a density of 2.70 g/cm³, what is the mass, in grams, of the foil?

Answers

Answer:

Mass of aluminium foil = 387.57_g

Explanation:

density of aluminium foil 2.70 g/cm³,

1_yd = 91.44_cm

1_in = 2.54_cm

Length of aluminum foil = 63.2_yd = 5779.008_cm

Width of aluminium foil = 11 in = 27.94_cm

Thickness of aluminium foil = 0.00035_in. = 0.000889_cm

Volume of aluminium foil = length × width × thickness = 5779.008_cm × 27.94_cm × 0.000889_cm = 143.54_cm^3

Mass of aluminium foil = (volume of aluminium foil) × (density of aluminium foil) = 143.54_cm³ × 2.70 g/cm³ = 387.57_g

Final answer:

To find the mass of the aluminium foil, calculate the volume and multiply it by the density. The mass of the foil is 0.3878 grams.

Explanation:

To find the mass of the aluminium foil, we need to calculate the volume and then multiply it by the density. First, let's convert the dimensions to the same unit. The aluminium foil is 63.2 yd long, which is equivalent to 190.4 ft. The width is 11 in, which is 0.92 ft, and the thickness is 0.00035 in, which is 2.92e-5 ft.

The volume of the foil can be calculated by multiplying the length, width, and thickness.

V = (190.4 ft) * (0.92 ft) * (2.92e-5 ft).

This gives us a volume of 5.0634e-3 cubic feet.

Next, we need to convert the volume to cubic centimetres because the density is given in g/cm³.

There are 28.3168 cubic centimetres in one cubic foot, so the volume in cm³ is

5.0634e-3 * 28.3168 = 0.1433 cm³.

Finally, we can calculate the mass by multiplying the volume by the density.

Mass = 0.1433 cm³ * 2.70 g/cm³ = 0.3878 grams.

So, The mass of the foil is 0.3878 grams.

Learn more about Mass of aluminium foil here:

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A 1170.-gram sample of NaCl() completely reacts, producing 460. grams of Na(). What is the total mass of Cl2(g) produced?

Answers

2 NaCl --------> 2 Na + Cl₂

2 mol * 23 g Na = 46 g

2*35,5 g Cl = 71

46 g Na --------- 71 g Cl₂
460 g Na -------- ?

mass ( Cl₂) = 460 . 71 / 46

Mass (Cl₂) = 32660 / 46

m= 710 g of Cl₂

hope thips helps

Hydrogen bonding is a type of(1) strong covalent bond
(2) weak ionic bond
(3) strong intermolecular force
(4) weak intermolecular force

Answers

Option 3. strong intermolecular force

Certain substances such as H2O, HF, NH3 from hydrogen bonds, and the formation of which affects properties (mp,bp,solubility) of substance. Other compounds containing OH and NH2 groups also form hydrogen bond. Molecules of many organic compounds such as alcohols, acids, amines, and amino acids contain these groups, and thus hydrogen bonding plays an important role in biological science.

Calculate ℰ° values for the galvanic cells described below. (a) cr3+(aq) + cl2(g) equilibrium reaction arrow cr2o72-(aq) + cl -(aq) v (b) io3-(aq) + fe2+(aq) equilibrium reaction arrow fe3+(aq) + i2(aq)

Answers

Answer:

$\boxed{\text{(a) 0.00 V; (b) 0.424 V}}$

Explanation:

We must look up the standard reduction potentials for the half-reactions.

ℰ°

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

ℰ°/V

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00

(b) Fe²⁺/IO₃⁻

ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424

The ℰ° values for the cells are $\boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}$

Answer:

Answer:

\boxed{\text{(a) 0.00 V; (b) 0.424 V}}

Explanation:

We must look up the standard reduction potentials for the half-reactions.

ℰ°

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

ℰ°/V

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00

(b) Fe²⁺/IO₃⁻

ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424

The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}

Explanation:

its right trust

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