List some of applications with chemistry in daily life

Help me please im in 4th grade and online is frustrating

Snowcat [4.5K]

Answer:

A

Explanation:

The answer is A) because the A container has tightly packed molecules but not as tight as C so that means A is liquid, and C is a solid which makes B a gas

7 0

3 years ago

What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

Round

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.4184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately 147 grams.

3 0

3 years ago

ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea

kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

Step 1: Data given

H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Step 2: Calculate ΔG

ΔG = ΔG° + RTln Q

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

4 0

3 years ago

A+common+iv+solution+is+0.9%+saline+(nacl+solution).+what+is+the+osmolarity+of+0.9%+saline+mosmoles/l?+the+molecular+weight+of+n

Vedmedyk [2.9K]

An osmolarity of saline solution is 308 mosmol/L.

m(NaCl) = 9 g; the mass of sodium chloride

V(solution) = 1 L; the volume of the saline solution

n(NaCl) = 9 g ÷ 58.44 g/mol

n(NaCl) = 0.155 mol; the amount of sodium chloride

number of ions = 2

Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.

The osmolarity = n(NaCl) ÷ V(solution) × 2

The osmolarity = 0.154 mol ÷ 1 L × 2

The osmolarity = 0.154 mol/L × 1000 mmol/m × 2

The osmolarity of the saline solution = 308 mosm/L.

More about osmolarity: brainly.com/question/13258879

#SPJ4

8 0

1 year ago

rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o

SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

Initial important note:

Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.

1) Air:

Source: internet

Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.

2) Exhaled air:

Source: internet.

Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂

3) Air from the decomposition of H₂O₂.

In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:

i) Chemical Equation:

H₂O₂ (g) → H₂ (g) + O₂ (g)

ii) mole ratio of the products 1 mol H₂ : 1 mol O₂

iii) convert moles into mass (grams)

1 mol H₂ × 2 × 1.008 g/mol = 2.016 g

1 mol O₂ × 2 × 15.999 g/mol = 31.998 g

Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%

4) Air from the decomposition of NaHCO₃:

i) chemical equation:

2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)

ii) mole ratio: take into account only the gases in the products:

1 mol CO₂ (g) : 1 mol H₂O

iii) mass in grams

CO₂: molar mass ia approximately 44.01 g/mol

H₂O: molar mass is approximately 18.02 g/mol

iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.

5) The result is:

H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

7 0

3 years ago

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List some of applications with chemistry in daily life​

List some of applications with chemistry in daily life