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3 years ago

How does the atmosphere make conditions on earth suitable for living things

1 answer:

8 0

The atmosphere provides C02 for us, it protects us from the Sun's UV rays, it protects us from many objects in space that could crash into earth without it, and it holds the moisture in the air.

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Ppredict the identity of the precipitate in the below reaction: BaCl2(aq) + K2SO4(aq) →

Annette [7]

Answer: The precipitate formed is $BaSO_4$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid or precipitated form are represented by (s) after their chemical formulas.

A double displacement reaction in which one of the product is formed as a solid is called as precipitation reaction.

The balanced chemical equation is:

$K_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2KCl(aq)$

3 0

3 years ago

A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = $\frac{18.0 g}{60 g/mol}=0.3 mol$

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

$Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}$

Molarity of the urea solution ;

$M=\frac{0.3 mol}{0.200 L}=1.50 M$

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

$m=d\times V=0.95 g/mL\times 200.0 mL=190 g$

$m = 190 g = 190 \times 0.001 kg = 0.19 kg$

(1 g = 0.001 kg)

Molality of the urea solution ;

$Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

$m=\frac{0.3 mol}{0.19 kg}=1.58 m$

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0

3 years ago

Standard atmospheric pressure corresponds to the normal air pressure at sea level.

jekas [21]

Im not really sure what your asking.... Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, 14.70 pounds per square inch, 1,013.25 × 10 3 dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.

""atmospheric pressure | Britannica.com""

6 0

3 years ago

A student calculated the density of a sample of graphite to be 2.3 g/cm3. Show a numerical setup for calculating the student’s p

LUCKY_DIMON [66]

Answer:

Percent error = 1.5%

Explanation:

Given data:

Measured value of density of graphite = 2.3 g/cm³

Percent error = ?

Solution:

Formula:

Percent error = [Measured value - Actual value / actual value] × 100

Actual/accepted value of density of graphite = 2.266 g/cm³

Now we will put the values:

Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100

Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100

Percent error = 0.015 × 100

Percent error = 1.5%

8 0

3 years ago

Read 2 more answers

Number of stable isotopes in Hydrogen?

kari74 [83]

3.
protium (A = 1), deuterium (A = 2), and tritium (A = 3).

5 0

2 years ago

Read 2 more answers