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4 years ago

Sugar dissolves readily in water because it is a(n) ____ substance.

1 answer:

3 0

The answer to this question would be D. Hydrophilic.

The word hydrophilic mean attracted by water. That means the molecule has a force to attract water molecule, thus be able to dissolve in water. The polarity of the molecule would determine whether a molecule hydrophilic or not.
Its opposite would be hydrophobic which the molecule can't dissolve in water. One example of this would be oil or fat. That is why sometimes it is called lipophilic too.

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a hydrate of cocl2 with a mass of 6.00 g is heated strongly. after cooling the mass of the a hydrate is 3.27g. how many grams of

nlexa [21]

The term hydrate would mean that the compound contains water molecules. When this compound is heated, the water molecules are lost first which means that what remains would be the dried compound. Therefore, the H2O lost from the hydrate would be 6.00g - 3.27g = 2.73 g.

6 0

4 years ago

Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B

Ivenika [448]

Answer:

-2546 kJ

Explanation:

It is possible to obtain the enthalpy of a reaction from the sum of different intermediate reactions.

For the reaction:

4B(s) + 3O₂(g) → 2B₂O₃(s)

The intermediate reactions are:

A- B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g), ΔH°A= +2035 kJ

B- 2B(s) + 3H₂(g) → B₂H₆(g), ΔH°B= +36 kJ

C- H₂(g) + 1/2O₂(g) → H₂O(l), ΔH°C= -285 kJ

D- H₂O(l) → H₂O(g), ΔH°D= +44 kJ

2B = 4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH°2B= +78 kJ

-2A = 6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH°-2A= -4070 kJ

-6C = 6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH°-6C= +1710 kJ

-6D = 6H₂O(g) → 6H₂O(l) ΔH°-6D = -264 kJ

The sum of 2B - 2A - 6C - 6D produce:

4B(s) + 3O₂(g) → 2B₂O₃(s)

And the enthalpy is: ΔH°2B + ΔH°-2A + ΔH°-6C + ΔH°-6D = <em>-2546 kJ</em>

I hope it helps!

3 0

3 years ago

A variable that is not being directly tested during an experiment should be ____.

makkiz [27]

A variable that is not being directly tested during an experiment should be independent

7 0

4 years ago

Matter is anything that has both mass and ____

mrs_skeptik [129]

Answer:

volume

Explanation:

6 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Solving the Rydberg equation for energy change gives ΔE = R[infinity]hc [ 1 n12 − 1 n22 ] where the

icang [17]

Answer:

Explanation:

Utilizing Rydber's equation:

ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have

n=1 to n= infinity

ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)

n= 3 to n= infinity

ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J

b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .

1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹

1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m

λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm

7 0

3 years ago

Read 2 more answers