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Ostrovityanka [42]
3 years ago
6

Why is the Hubble telescope located in space? *

Chemistry
1 answer:
dolphi86 [110]3 years ago
7 0

Answer: I’m not one hundred percent sure, but based off of what I know, I believe it is most likely “So it’s images aren’t distorted by the Earth’s atmosphere.”

Explanation:

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True or False;<br> aa is faster moving lava than pahoehoe?
frosja888 [35]

Answer:

It is false :l

Explanation:

Pahoehoe is a smooth and continuous lava crust. Pahoehoe forms when the effusion rate is low and consequently the velocity of lava flow is slow. Pahoehoe lava flow is usually at least 10 times slower than typical aa lava flow.

:/

7 0
3 years ago
What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?
Vadim26 [7]

Answer:

158 L.

Explanation:

What is given?

Pressure (P) = 1 atm.

Temperature (T) = 112 °C + 273 = 385 K.

Mass of methane CH4 (g) = 80.0 g.

Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

PV=nRT.

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.

So, in this case, n=5.

Now, let's solve for 'V' and replace the given values in the ideal gas law equation:

V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}

The volume would be 158 L.

6 0
1 year ago
Calculate the volume of 0.10 g of titanium (4.51 g/cm³).
Juli2301 [7.4K]

The volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

<h3>What is volume?</h3>

Volume is known to be equal to the mass divided by the density.

It is written thus:

Volume = Mass / density

<h3>How to calculate the volume</h3>

The volume is calculated using the formula:

Volume = mass ÷ density

Given the mass = 0. 10g

Density = 4.51 g/cm³

Substitute the values into the formula

Volume of titanium = 0. 10 ÷ 4.51 = 0. 02 cm³

Thus, the volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

Learn more about volume here:

brainly.com/question/1762479

#SPJ1

4 0
2 years ago
Classify the below solids as amorphous or crystalline. emerald glass MgCl, rubber​
svet-max [94.6K]

Answer:

Here yo

Explanation:

6 0
2 years ago
Read 2 more answers
20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as
shepuryov [24]

The reaction forms 98.76 g AlCl_3.  

We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1. Gather all the information</em> in one place with molar masses above the formulas and everything else below them.  

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

Mass/g: 20.00 _78.78

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

Moles of Al = 20.00 g Al × (1 mol Al /26.98 g Al) = 0.741 29 mol Al

Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of AlCl_3 we can obtain from each reactant.  

<em>From Al</em>: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

<em>From Cl_2</em>: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.

<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

4 0
3 years ago
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