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2 years ago

Compared to the density of liquid water, the density of an ice cube is

2 answers:

7 0

The ice structure takes up more volume than the liquid water molecules, hence ice is less dense than liquid water. What is the exact change in volume of the water when it freezes as ice? Example: Calculate the volume in a 100 g ice cube with a density of 0.92 g per mL.

arsen [322]2 years ago

6 0

Answer:

less than the water

Explanation:

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Which statement best describes frustration?

wlad13 [49]

C. when you can't achieve your goal due to events beyond your control

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2 years ago

If the mess of an object______, the weight of an object will _______

Gnom [1K]

Answer:

the same

Explanation:

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2 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

7 0

2 years ago

Name one metal and one non-metal from the periodic table? Y’all I need help Thank you!

Stella [2.4K]

Answer:

Zinc and Neon!

Explanation:

zinc is an element that is a transition metal and neon is a noble gas

8 0

2 years ago

Based on the replacement reaction, what would the products of the reaction be?

cricket20 [7]

Answer:

$\rm Be(OH)_2$ and $\rm (NH_4)_2 SO_4$ . The missing ion would be $\rm OH^{-}$ .

Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

$\rm BeSO_4$ , and
$\rm NH_4 OH$ .

In particular,

$\rm BeSO_4$ is made up of $\rm Be^{2+}$ ions and $\rm {SO_4}^{2-}$ ions, while
$\rm NH_4 OH$ is made up of $\rm {NH_4}^{+}$ ions and $\rm OH^{-}$ ions.

In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

$\rm Be^{2+}$ is a cation. In $\rm BeSO_4$ , $\rm Be^{2+}$ was bounded $\rm {SO_4}^{2-}$ anions. During the reaction, it bonds with $\rm OH^{-}$ anions to produce $\rm Be(OH)_2$ .

$\rm {NH_4}^{+}$ is also a cation. In $\rm NH_4 OH$ , $\rm {NH_4}^{+}$ was bounded to $\rm OH^{-}$ ions. During the reaction, it bonds with $\rm {SO_4}^{2-}$ anions to produce $\rm (NH_4)_2 SO_4$ .

Hence, the two products will be $\rm Be(OH)_2$ and $\rm (NH_4)_2 SO_4$ .

Note that charges on the ions must balance. For example, a $\rm Be^{2+}$ ion carries twice as much charge as an $\rm {NH_4}^{+}$ ion. As a result, each $\rm Be^{2+}$ ion would bond with twice as many $\rm OH^{-}$ ions as $\rm {NH_4}^{+}$ would in $\rm NH_4 OH$ .

4 0

3 years ago

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