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2 years ago

Please help!! Question on the picture

Chemistry

1 answer:

garik1379 [7]2 years ago

5 0

D) You need more information to figure this out.

The atom has three protons, so you know from the Periodic Table that the atom is lithium, Li.

However, there are two stable isotopes of Li.

The most common isotope has <em>four</em> neutrons, and the other one has three.

You have <em>no way of knowing</em> which isotope the diagram represents.

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Removing one electron from an atom results in the formation of an Question 7 options: ion with a 1- charge. ion with a 7+ charge

olga nikolaevna [1]

I believe it is an ion with a 1+ charge. If you remove an electron from an atom it will have a positive charge. But if you add electrons it will be a negative charge. Hope I helped!

5 0

3 years ago

How many moles of Cl- ions are needed to completely combine with 0.25 moles of Mg+2 ions?

Mnenie [13.5K]

<span>The number of moles of Cl- ions needed to combine completely with 0.25 mole of Mg+2 ions is:
0.50.</span>

4 0

2 years ago

How many moles of hydrogen are in 3. 06 × 10⁻³ g of glycine , c₂h₅no₂?.

Gala2k [10]

Answer:

n = 6.06 x $10^{-4}$ mol

Explanation:

n =?

m = 3.06 x 10-³ g

M (H5) = 5 x 1.01 (Since we only want hydrogen)

Atomic mass of C = 12.01

Atomic mass of H is 1,01, etc.

Having this data, we can use the Molar mass formula and change it so we can know the quantity of matter (n) in moles, and we just replace it.

M = $\frac{m}{n}$ ⇔ n = $\frac{m}{M}$ ⇔ n = $\frac{3.06 x 10^{-3} }{5,05}$ ⇔ n = 6.06 x $10^{-4}$ mol

Note: The numbers I've used may be different from yours, by a small difference. I don't know if it's the case, but hope it helped.

8 0

1 year 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

(b) What amount (mol) of H₂ can be produced from the given mass of H₂O?

Luden [163]

8 moles of H 2O are produced.

First, we need to figure out the chemical equation for producing water with oxygen which is H 2 + O2 = H 2O. Then, we need to balance the equation, resulting in 2H 2 + O2 = 2H 2O.

<h3>How many moles of H2 are required to make one mole of NH3?</h3>

Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.

<h3>How many moles of h2 are needed to produce a solution?</h3>

An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).

Learn more about H2O:

brainly.com/question/2193704

#SPJ4

8 0

1 year ago