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Andrei [34K]
3 years ago
12

My car has an internal volume of 3,600 L. If the sun heats my car from a temperature of 18.0 degrees Celsius to a temperature of

45.0 degrees Celsius, what will the pressure inside my car be? Assume the pressure a was initially 101.3 kPa.
Chemistry
1 answer:
Papessa [141]3 years ago
5 0

Answer:

110.7 kpa or 1.097 atm your choice of pressure units

Explanation:

We can assume the Volume is constant since we are not told other wise and it is indeed a car. We can Use P1/T1= P2/T2

So (T2xP1 )/ T1 = P2

(318Kx101.3kpa)/291k = 110.7 kpa or 1.097 atm your choice of pressure units

if your wondering why i used Kelvin gas laws are always in units of Kelvin

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What volume of nitrogen (n2) would be completely consumed in the reaction with 30.80 g of
Shtirlitz [24]

The answer is 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

<h3>What is a mole ?</h3>

A mole is defined as 6.02214076 × 10²³ atoms, molecules, ions, or other chemical units.

Write a balanced equation.

Calculate the moles of H₂ in 30.8 g.

Calculate the moles of N₂ required to react with H₂.

Calculate the mass of N₂.

Calculate the initial mass of N₂.

Start with a balanced equation.

N₂ + 3H₂ --> 2NH₃

Calculate the moles of H₂ in 30.8 g.

n = m/M; where n = moles, m = mass, and M = molar mass.

M(H₂) = 1.008 g/mol

n(H₂) = (30.8 g)/(1.008 g/mol) = 30.56 mol H₂

Calculate the moles of N₂ required to react with 30.56 mol H₂ , using the mole ratio between H₂ and N₂ in the balanced equation.

30.56 mol H₂ × 1 mol N₂/3 mol H₂ = 10.18 mol N₂

Calculate the mass of N₂ in 10.18 mol.

m = n × M

M(N₂) = 2 × 14.007 g/mol N = 28.014 g/mol N₂

m(N₂) = 10.18 mol × 28.014 g/mol = 285.33g N₂

Therefore 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

To know more about mole

brainly.com/question/26416088

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5 0
2 years ago
25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca
Kamila [148]

Answer:

Concentration of the original \rm KOH solution: approximately 1.05\; \rm mol \cdot dm^{-3}.

Explanation:

Notice that the concentration of the \rm HCl solution is in the unit \rm mol\cdot dm^{-3}. However, the unit of the two volumes is \rm cm^{3}. Convert the unit of the two volumes to \rm dm^{3} to match the unit of concentration.

\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}.

\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 0.0350\; \rm dm^{3} of 0.75\; \rm mol \cdot dm^{3} \rm HCl\! solution:

\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}.

\rm HCl is a monoprotic acid. In other words, each \rm HCl\! would release up to one proton \rm H^{+}.

On the other hand, \rm KOH is a monoprotic base. Each \rm KOH\! formula unit would react with up to one \rm H^{+}.

Hence, \rm HCl molecules and \rm KOH\! formula units would react at a one-to-one ratio.

{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l).

Therefore, that 0.02625\; \rm mol of \rm HCl molecules would neutralize exactly the same number of \rm NaOH formula units. That is: n(\mathrm{NaOH}) = 0.02625\; \rm mol.

Calculate the concentration of a \rm NaOH solution where V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3} and n(\mathrm{NaOH}) = 0.02625\; \rm mol:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}.

7 0
3 years ago
4.14 Oxygen requirement for growth on glycerol Klebsiella aerogenes is produced from glycerol in aerobic culture with ammonia as
Iteru [2.4K]

Answer:

0.37 g

Explanation:

The molecular weight for Glycerol = 92

Number of Carbon atoms in glycerol (x)  C_3H_8O_3 = 3

Molecular weight of  the Biomass  ( Klebsiella aerogenes )

= CH_{1.73}O_{0.43}N_{0.24}

= \frac{23.97}{0.92}

= 26.1

From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as \delta _g) as follows:

= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)

= 4.15

Given that the yield of the Biomass = 0.40 g

However;

C = Yield of Biomass *\frac{Molecular weight of substrate}{Molecular weight of the Biomass}

C = 0.40*\frac{92}{26.1}

C = 1.41 g

Now , the oxygen requirement can be calculated as:

= \frac{1}{4}*(n*S -  C * \delta _{g})

= \frac{1}{4}(3*4.7-1.41*4.15)

= 2.1 g/mol

Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.

Now converting it to mass terms; we have:

= 2.1*\frac{number of mole of oxygen}{molecular weight of glycerol}

= 2.1 * \frac{16}{92}

= 0.3652 g

≅ 0.37 g

∴ The oxygen requirement for this culture in mass terms = 0.37 g

3 0
3 years ago
What is the most likely mode of decay for each?(c) ¹¹⁰₄₁4Ru
zmey [24]

The most likely mode of decay for a neutron-rich nucleus is one that converts a neutron into a proton.

<h3>What is radioactive decay?</h3>

The process through which an unstable atomic nucleus loses energy through radiation is known as radioactive decay, also known as nuclear decay, radioactivity, radioactive disintegration, or nuclear disintegration. A substance that has unstable nuclei is regarded as radioactive. Alpha decay, beta decay, and gamma decay are three of the most frequent kinds of decay, and they all entail the emission of one or more particles. Beta decay is a result of the weak force, while the nuclear force and electromagnetism are in charge of the other two mechanisms. The capture of an inner electron from one of the electron shells by an unstable nucleus is the fourth prevalent form of decay.

To learn more about radioactive decay visit:

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6 0
2 years ago
5 lists of physical change
Alecsey [184]
Crushing a can, breaking glass, cutting paper, boiling water, chopping wood, and mixing water and sand
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