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Ahat [919]
2 years ago
11

Jean-Paul is racing in the Tour de France bicycle race. He is moving along rapidly at a speed of 15 kilometers per hour. Suddenl

y, a strong wind begins to blow into his face. Jean-Paul cannot exert any more force on the pedals. What will happen to Jean-Paul's speed while the wind continues to blow against him?
Chemistry
1 answer:
Murrr4er [49]2 years ago
5 0
He’ll slow down because the wind is blowing into his face
You might be interested in
A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0
3 years ago
An isotope of cesium-137 has a half-life of 30 years. If 5.0g of cesium-137 decays over 60 years. How many grams will remain.
LenKa [72]

Answer:

1.25 gram of cesium-137 will remain.

Explanation:

Given data:

Half life of cesium-137 = 30 year

Mass of cesium-137 = 5.0 g

Mass remain after 60 years = ?

Solution:

Number of half lives passed = Time elapsed / half life

Number of half lives passed = 60 year / 30 year

Number of half lives passed = 2

At time zero = 5.0 g

At first half life = 5.0 g/2 = 2.5 g

At 2nd half life = 2.5 g/ 2 = 1.25 g

Thus. 1.25 gram of cesium-137 will remain.

6 0
2 years ago
What is the concentration in milligrams per milliter of a solution containing 23.5 meq sodium chlorise per milliliter? mw nacl =
ValentinkaMS [17]

1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.

It is calculated in mg/ml.

The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.

Formula for calculating concentration in mg/ml is

Conc. (mg/ml) = M(eq) /ml ×  Molecular weight / Valency

Given

M(eq) NaCl/ ml = 23.5

Molecular weight pf NaCl = 58.5 g/mol

Valency = 1

Putting the values into the formula

Conc. (mg/ml) = 23.5 ×58.5/1

                       = 1374.75 mg/ml

Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Learn more about Concentration here brainly.com/question/14500335

#SPJ4

7 0
2 years ago
Which of the following molecules has one lone pair of electrons?
BigorU [14]
You can automatically rule out CH₄ since it has no lone pairs at all around the central atom. Water has 2. Ammonia is the only Lewis structure that contains one lone pair.
7 0
2 years ago
Read 2 more answers
How many atoms are in 2.45 moles of hydrogen
allochka39001 [22]

Answer:

There are 1.4754246675000002e+24 atoms of Hydrogen within the measurement of 2.45 moles of hydrogen!

Explanation:

6 0
3 years ago
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