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4 years ago

14

A rock dropped in a graduated cylinder raises the level of water from 20 mL to 35 mL. The rock has a mass of 45 g. What is the d

ensity of the rock?

1 answer:

Brut [27]4 years ago

7 0

I think 15 I might be wrong

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Corrosion is what type of chemical change?

Nostrana [21]

Answer A

Oxidation.

hope this helps!

5 0

3 years ago

How many grams are in 3.93 x 10^24 molecules of CCl4?

Lorico [155]

<h3>Answer:</h3>

1000 g CCl₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.93 × 10²⁴ molecules CCl₄

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.93 \cdot 10^{24} \ molecules \ CCl_4(\frac{1 \ mol CCl_4}{6.022 \cdot 10^{23} \ molecules \ CCl_4})(\frac{153.81 \ g \ CCl_4}{1 \ mol \ CCl_4})$
Multiply: $\displaystyle 1003.77 \ g \ CCl_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1003.77 g CCl₄ ≈ 1000 g CCl₄

8 0

3 years ago

If the initial amount of iridium-192 produces radiation at 1,000 times the safe level, how long will it take before radiation le

Contact [7]

To cause death within hours of exposure to radiation, the dose needs to be very high, 10Gy or higher, while 4-5Gy will kill within 60 days, and less than 1.5-2Gy will not be lethal in the short term. However all doses, no matter how small, carry a finite risk of cancer and other diseases. Patients exposed to radiation between 8 to 30 Gy experience nausea and severe diarrhea within an hour, and they die between 2 days and 2 weeks after exposure. Absorbed doses greater than 30 Gy cause neurological damage

4 0

3 years ago

In an experiment, zinc chlorate decomposed according to the following chemical equation.

mote1985 [20]

Answer : The correct expression is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Explanation :

First we have to calculate the moles of $Zn(ClO_3)_2$

$\text{ Moles of }Zn(ClO_3)_2=\frac{\text{ Mass of }Zn(ClO_3)_2}{\text{ Molar mass of }Zn(ClO_3)_2}=\frac{150g}{232.29g/mole}=\frac{150}{232.29}mole$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$Zn(ClO_3)_2\rightarrow ZnCl_2+3O_2$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Zn(ClO_3)_2$ react to give 3 mole of $O_2$

So, $\frac{150}{232.29}$ moles of $Zn(ClO_3)_2$ react to give $\frac{150}{232.29}\times \frac{3}{1}$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(\frac{150}{232.29}\times \frac{3}{1})mole\times (31.998g/mole)=\frac{150\times 3\times 31.998}{232.29\times 1}g$

Therefore, the correct expression for the mass of oxygen gas formed is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

6 0

3 years ago

Read 2 more answers

Perform the following conversions:

ololo11 [35]

Answer:

a) $20^0C$ and $293K$

b) $437K$ and $327.2^0F$

c) $-273^0C$ and $459.44^0F$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ , $^0F$ and $K$

These units of temperature are inter convertible.

$t^0C=(t+273)K$

$t^oC=\frac{5}{9}\times (t^oF-32)$

$K-273=\frac{5}{9}\times (^oF-32)$

a) 68°F (a pleasant spring day) to °C and K.

Converting this unit of temperature into $^0C$ and $K$ by using conversion factor:

$t^oC=\frac{5}{9}\times (68^oF-32)$

$t=20^0C$

$20^0C=(20+273)K=293K$

b) 164°C (the boiling point of methane, the main component of natural gas) to K and °F

Conversion from degree Celsius to Kelvins and Fahrenheit

$164C=\frac{5}{9}\times (t^oF-32)$

$t^0F=327.2$

$164°C=(164+273)K=437 K$

c) 0K (absolute zero, theoretically the coldest possible temperature) to °C and °F.

$K-273=\frac{5}{9}\times (t^oF-32)$

$0-273=\frac{5}{9}\times (t^oF-32)$

$t=-459.4^0F$

$t^K=(0-273)^0C=-273^0C$

8 0

3 years ago