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3 years ago

7

What is the density of ice

2 answers:

Nadusha1986 [10]3 years ago

8 0

0.9167 g/cm3<span>The density of ice is <span>0.9167 g/cm3</span> at 0 °C, whereas water has a density of 0.9998 g/cm³ at the same temperature. Liquid water is densest, essentially 1.00 g/cm³, at 4 °C and becomes less dense as the water molecules begin to form the hexagonal crystals of ice as the freezing point is reached.</span>

amm18123 years ago

5 0

The density of ice is 0.9167 g/cm<span>3</span>

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Why are animal vaccinated?

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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

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2 years ago

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3 years ago

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How heavy is one atom of hydrogen

Harman [31]

Hello!

the atomic mass of one atom of hydrogen is equal to 1.00784 in atomic mass units

one atomic mass unit is equal to 3.6609e-27 pounds

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3 years ago

Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromi

GuDViN [60]

<u>Answer:</u> The mass of bromine reacted is 160.6 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 18.1 g

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Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol$

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3 years ago