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2 years ago

PLZ HELP ASAP

Chemistry

1 answer:

Sidana [21]2 years ago

3 0

Answer:

plant cells and animal cell have similaritys but also haev diffrences.

Explanation:

one example of a diffrence bettwen them is that plants have a cell wall but a animal doesnt even tho they are both living organizoums

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The formulas for elements never contain a subscript true or not true

Andru [333]

The answer should be false. Elements contain only one atom.

7 0

3 years ago

What is the molarity of a solution in which 55. 49 g of calcium chloride is dissolved in enough water to make 500. ml of solutio

PSYCHO15rus [73]

The molarity of a solution in which 55. 49 g of calcium chloride is dissolved in enough water to make 500. ml of solution is 1M.

<h3>What is molarity? </h3>

It is defined as number of moles of solute divided by volume of solution.

Given,

Mass of CaCl2 =55.49g

Molar mass of CaCl2 =40+35+35=110g

Mole= given mass/ molar mass

= 55.49/110=0.50mol.

Now, putting all values we get the molarity

Molarity =0.5×1000/500=1M

Thus, the molarity of given solution is 1M.

learn more about Molarity:

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4 0

1 year ago

If 56 grams of notrogen are used up by the reaction, how many grams of amonia will be produced? 1N2+3H2--> 2NH3

djverab [1.8K]

1 mole of N2 produces 2 moles of NH3
OR...
14 x 2 grams of N2 produces 2(14 +3) grams of NH3
1 gram of N2 produces 34/28 grams of NH3
therefore, 56 grams produce (34/28 )x 56 =68 grams of NH3

the answer thus would be 68 grams of NH3

5 0

3 years ago

Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago

Would any Chemistry braniacs be kind enough to help explain how to work these out? Any help is appreciated, thanks

o-na [289]

Answer:

NO.3) Mass of Al2O3 formed = 229.5g

7 0

3 years ago