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erastovalidia [21]
3 years ago
11

Please complete the sentence. Matter can be classified as a(n) ____________ or a(n) ________________?

Chemistry
2 answers:
Damm [24]3 years ago
7 0

substance or a mixture

butalik [34]3 years ago
4 0

Matter can be classified as a(n) substance or a(n) mixture.


Substance:

Pure, only atoms of the same element


Mixture:

Not pure, atoms of different elements mixed together.



Hope it helped,


BioTeacher101

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Some antacid tables contain aluminum hydroxide. The aluminum hydroxide reacts with stomach acid according to the equation: Al(OH
MatroZZZ [7]

Answer:

26.67 mol HCl

Explanation:

Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O

In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.

To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:

  • 8.89 mol Al(OH)₃ * \frac{3molHCl}{1molAl(OH)_{3}} = 26.67 mol HCl

Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.

4 0
3 years ago
What are some visible signs of an acid-base reaction?
marta [7]
I think one of the signs is <span>water and  salt are formed 

</span>
6 0
3 years ago
How many grams are in 5.2 moles of Li2SO4
skelet666 [1.2K]

Answer:

572 g

Explanation:

Molar mass is the mass of 1 mol of an element or compound

molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound

molar masses of each element making up lithium sulphate

Li - 7 g/mol

S - 32 g/mol

O - 16 g/mol

molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )

molar mass = 110 g/mol

mass of 1 mol of Li₂SO₄ is 110 g

therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g

mass is 572 g

7 0
3 years ago
Argon has a pressure of 34.6 atm. It is transferred to a new tank with a volume of 456 L and pressure of 2.94 atm. What was the
NemiM [27]

Answer:

38.75 L

Explanation:

From the question,

Applying Boyles Law,

PV = P'V'....................... Equation 1

Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' =  Final Volume of Argon gas.

make V the subject of the equation

V = P'V'/P.................... Equation 2

Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.

Substitute these values into equation 2

V = (456×2.94)/34.6

V = 38.75 L

3 0
3 years ago
Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calcu
stealth61 [152]

Answer:

Explanation:

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

Explanation;

According to Arrhenius equation:

i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K

T2 = 25 oC = (25 + 273) K = 298 K

i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)

i.e. ln(k2/0.000717) = -2.54738

i.e. k2/0.000717 = e^{-2.54738}

= 0.078286

Therefore, the required constant (k2) = 0.078286 * 0.000717 = 5.61*10^-^5

6 0
3 years ago
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