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3 years ago

What happens to temperature as objects are heated up and cooled down?

1 answer:

8 0

If heat is transferred from an object to the surroundings, then the object can cool down and the surroundings can warm up. When heat is transferred to an object by its surroundings, then the object can warm up and the surroundings can cool down.

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NEED HELP FAST!!

koban [17]

Answer:

I'm pretty sure that's right.

Explanation:

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8 0

3 years ago

If a volume of gas 177.6mL was collected at a temperature of 25.8C and the pressure is 799.7 torr, what is the original concentr

Step2247 [10]

Answer:

The original concentration of the acid was 0.605 M

Explanation:

Step 1: Data given

Volume of gas = 177.6 mL = 0.1176 L

Temperature = 25.8 °C = 298.95 K

Pressure = 799.7 torr = 799.7/ 760 = 1.0522368 atm

Volume of acid needed to react = 12.6 mL = 0.0126 L

Step 2: Calculate moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles = TO BE DETERMINED

⇒with p = the pressure of the gas = 799.7 torr = 1.0522368 atm

⇒with V = the volume of the gas = 177.6 mL = 0.1776 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 25.8 °C = 298.95 K

n = 0.007618 moles

Step 3: Calculate original concentration

We need 0.007618 moles of acid to react with the same amount of moles gas

Concentration acid = moles / volume

Concentration acid = 0.007618 moles / 0.0126 L

Concentration acid = 0.605 M

The original concentration of the acid was 0.605 M

5 0

3 years ago

If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)

Ludmilka [50]

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

$w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%$

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

4 0

3 years ago

Formula semidesarrollada de los hidrocarburos aromáticos de 1 a 10 átomos de carbono

Tju [1.3M]

Answer:

El principal componente del gas natural es también el hidrocarburo más simple: el metano. Este compuesto está formado por un átomo de carbono y cuatro átomos de hidrógeno y se representa de dos formas:

El hidrocarburo que le sigue en simplicidad es aquel que está constituido por dos átomos de carbono. Su fórmula condensada es C2H6 y se le conoce como etano.

Si se continúan colocando átomos de carbono con enlaces sencillos entre ellos e hidrógenos en los enlaces libres, se crean largas cadenas de compuestos. Al etano le sigue el propano (C2H8) y a éste, el butano (C4H10). Todos estos compuestos forman parte de la familia de los alcanos, y sus nombres terminan con el sufijo –ano para indicar que pertenecen a la misma familia.

3 0

3 years ago

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

5 0

3 years ago