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3 years ago

A substance that is present in a reaction but not used up in the reaction is called a _____.

Chemistry

2 answers:

Goryan [66]3 years ago

7 0

A substance that is present in a reaction but not used up in the reaction is called a catalyst

Blababa [14]3 years ago

4 0

Answer: A. Catalyst

Explanation: A chemical reaction is a reaction between reactants which, due to different chemical processes during the reaction, yields a particular reaction product. A catalyst is a compound or substance that does not participate in the process as a reactant, therefore it does not affect the chemical composition of the compound obtained by the reaction, but only serves to speed up the chemical process. So the catalyst is not part of the chemical process as a reactant or product, just as a mechanism for faster reaction.

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A solution containing 4.15 g of magnesium chloride is added to a second solution containing excess

AveGali [126]

Answer:

1. 5.24 g

2. 75.4%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

MgCl₂ + Na₂SO₄ —> MgSO₄ + 2NaCl

Next, we shall determine the mass of MgCl₂ that reacted and the mass of MgSO₄ produced from the balanced equation. This can be obtained as follow:

Molar mass of MgCl₂ = 24 + (35.5×2)

= 24 + 71

= 95 g/mol

Mass of MgCl₂ from the balanced equation = 1 × 95 = 95 g

Molar mass of MgSO₄ = 24 + 32 + (16×4)

= 24 +32 + 64

= 120 g/mol

Mass of MgSO₄ from the balanced equation = 1 × 120 = 120 g

SUMMARY:

From the balanced equation above,

95 g of MgCl₂ reacted to produce 120 g of MgSO₄.

1. Determination of the ideal yield of magnesium sulfate, MgSO₄.

From the balanced equation above,

95 g of MgCl₂ reacted to produce 120 g of MgSO₄.

Therefore, 4.15 g of MgCl₂ will react to produce = (415 × 120)/95 = 5.24 g of MgSO₄.

Thus, the ideal yield of magnesium sulfate, MgSO₄ is 5.24 g

2. Determination of the percentage yield.

Actual yield of MgSO₄ = 3.95 g

Ideal yield of MgSO₄ = 5.24 g

Percentage yield =?

Percentage yield = Actual yield / Ideal yield × 100

Percentage yield = 3.95 / 5.24 × 100

Percentage yield = 395 / 5.24

Percentage yield of MgSO₄ = 75.4%

7 0

3 years ago

If you lose half of your mass, its gravity will.. causing the tides to ..

babymother [125]

Muchas gracias amigo

5 0

3 years ago

a weather balloon is inflated to a volume 2.2 10square3 L with 374g of helium. what is the density of helium in grams per liter

Kruka [31]

Answer:

Density = 0.17 g/L

Explanation:

It is given that,

Volume of the inflated balloon filled with Helium, $V=2.2\times 10^3\ L$

Mass, m = 374 g

We need to find the density of helium. It is equal to its mass per unit volume. It can be given by :

d =m/V

$d=\dfrac{374\ g}{2.2\times 10^3\ L}\\\\=0.17\ g/L$

So, the density of helium in the balloon is 0.17 g/L.

5 0

3 years ago

As the temperature of a sample of a radioactive element decreases, the half-life will

ki77a [65]

As the temperature of a sample of a radioactive element decreases, the half-life will remain the same

<u>Explanation:</u>

Half-life is defined as the time needed for half of the opening material to modify or decay. Half-lives can be estimated from measures on the variation in the mass of a nuclide and the period it gets to occur. In the period of that substance's half-life, half of the primary nuclei will decompose.

Although chemical changes were raced up or delayed down by varying factors such as temperature, concentration, etc, these factors do not affect half-life. Each radioactive isotope will have its unique half-life that is independent of any of these factors.

5 0

3 years ago

What is the total mass of NaNO3 that must be added to 70g in 100 grams of H2O to make a saturated solution at 30°C ?

dimulka [17.4K]

Answer:

58g

Explanation:

In order to solve this problem, you must take a look at the solubility graph for potassium nitrate.

Now, the solubility graph shows you how much solute can be dissolved per 100g of water in order to make an unsaturated, a saturated, or a supersaturated solution.

You're looking to make a saturated potassium nitrate solution using

50g of water at 60∘C. Your starting point will be to determine how much potassium nitrate can be dissolved in 100g of water at that temperature in order to have a saturated solution.

As you can see, the curve itself represents saturation.

If you draw a vertical line that corresponds to 60∘C and extend it until it intersects the curve, then draw a horizontal line that connects to the vertical axis, you will find that potassium has a solubility of about

115g per 100g of water. Your answer is 58g of potassium nitrate

7 0

4 years ago