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babunello [35]
3 years ago
12

How many grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl? Show all of the work needed to solve this

problem. Mg (s) + 2HCl (aq) yields MgCl2 (aq) + H2 (g)
Chemistry
1 answer:
Temka [501]3 years ago
8 0
Mghcl14 + H2 = mghclh 16 work needed to solve this problem
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Dan bikes 10 km west and then bikes another 5 km west. What is Dan's
777dan777 [17]
C

This is because 10+5=15
15/45=0.3
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2 years ago
This protein takes oxygen from hemoglobin in order to
never [62]

Answer:

i'm pretty sure it's 'transport oxygen from the lungs to tissues around the body'

4 0
3 years ago
Read 2 more answers
How many grams N2F4 can be produced from 225 g F,?​
zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

          2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

          2F₂     +    N₂   ----------->  N₂F₄

        2 mole                              1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

                 2F₂          +       N₂   ----------->  N₂F₄

        2 mole (38 g/mol)                        1 mole (104 g/mol)

                 76 g                                           104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

                  76 g of F₂ ≅ 104 g of N₂F₄

                   225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

                  X of N₂F₄ = 104 g x 225 g / 76 g

                  X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

7 0
3 years ago
Need help for number 4. do i use the q=mc DELTA T equation or the q=mHfusion?
Natasha2012 [34]
Since it is a phase change, use the mHfusion equation.
7 0
3 years ago
What is the concentration of an naoh solution that requires 15.0 ml of a 0.750 m h2so4 solution to neutralize 17.5 ml of naoh?
e-lub [12.9K]

So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
                      2NaOH  +  H₂SO₄   →  Na₂SO₄<span>  +  2H₂O
</span>
So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)

If 1000 ml  of H₂SO₄ contain 0.750 mol   [0.750 M is the amount of moles in 
                                                                       1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol     [15 ml is the amount of the acid that                                                                                  took part in the reaction]
    
⇒  x  =  \frac{15ml    *    0.750 mol}{1000 ml}
       
         = 0.01125 mol

Mole ratio of NaOH  to  H₂SO₄  can be obtained from the balanced equation
                 0 2NaOH  +  1H₂SO₄   →  Na₂SO₄  +  2H₂O

    mole ratio of   NaOH  to  H₂SO₄  is 2 : 1

∴ if mole of of H₂SO₄   =  0.01125 mol
   then moles of NaOH = (0.01125 mol) × 2
                                      = 0.0225 mol

If 17.5 ml of NaOH contain 0.0225 mol      [this was given in the question]
then let 1000 ml of NaOH contain  x

⇒ x  =  \frac{1000ml   * 0.0225 mol}{17.5 ml}
       
       = 1.286 mol

∴ concentration of NaOH is 1.286 mol/L 

8 0
3 years ago
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