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3 years ago

A taxi hurries with a costant speed of 84km/h. How far it travel in 5 hour

1 answer:

8 0

Hey there! If the taxi travels 84km in one hour (without making any changes in speed aka remaining constant), then multiply 84(5) to find how many kilometers it would drive in one hour. 84(5)=420km

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235 92 U >144 56 Ba + 3 1 on

Kobotan [32]

Answer:?

Explanation:

5 0

2 years ago

You combine 13 g of magnesium with 5 g of nitrogen to form a compound.

guapka [62]

Answer:

$\large \boxed{28 \, \% }$

Explanation:

magnesium + nitrogen ⟶ Product

13 g 5 g

Mass of product = 13 g + 5 g = 18 g

The product contains 5 g of nitrogen .

$\text{Percent N} = \dfrac{\text{5 g}}{\text{18 g}} \times \, 100\% = \mathbf{28 \, \%}\\\text{The percent by mass of nitrogen is $\large \boxed{\mathbf{28 \, \% }}$}$

7 0

3 years ago

Consider the formation of nitrogen dioxide from nitric

Temka [501]

Answer:

9 L

Explanation:

According to the question , the given reaction is -

2NO(g) + O₂(g)------->2NO₂(g)

Since ,

At STP ,

One mole of a gas occupies the volume of 22.4 L.

Hence , as given in the question -

9 L of NO , i.e .

22.4 L = 1 mol

1 L = 1 / 22.4 mol

9 L = 1 / 22.4 * 9 L = 0.40 mol

From the chemical reaction ,

The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .

Hence ,

2 moles of NO will produce 2 moles of NO₂.

Therefore ,

0.40 mol of NO will produce 0.40 mol of NO₂.

Hence , the volume of NO₂ can be calculated as -

1 mol = 22.4 L

0.40 mol = 0.40 * 22.4 L = 9 L

6 0

3 years ago

C12H7Cl3FNaO2 what are them elements in that chemical formula

LuckyWell [14K]

Assuming you are asking for the names of the elements in that formula , the answer is

carbon
hydrogen
chlorine
fluorine
sodium
oxygen

8 0

3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

mass N and mass percent:

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

2 years ago