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vovikov84 [41]
3 years ago
6

If Maria winks exactly 5 times every minute and she sleeps

Chemistry
2 answers:
Marianna [84]3 years ago
6 0

Answer:

4800 times does Maria wink every  day.

Explanation:

Number of time Maria winks= 5 times in minute = 5 winks/min

Number of hours she sleeps in day = 8 hours

1 day = 24 hours

Number of hours when Maria is not sleeping = 24 hour - 8 hour = 16 hours

1 hour = 60 minutes

16 hours = 16 × 60 min = 960 min

number of winks in 960  minutes:

960 min\times 5 winks/min=4800 winks

4800 times does Maria wink every  day.

Elina [12.6K]3 years ago
4 0

First, we need to find the amount of winks in an hour, to simplify it so we can answer easier. Since there are 60 minutes in an hour, we can multiply 5 by 60 to find the number of winks in an hour.

5*60=300

Now multiply 300 by 8.

300*8=2,400

She winks exactly 2,400 times per day.

Hope this helps!

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a student determined that it requires 106220 j of energy to vaporize 47g of water. is the student is right​
shusha [124]
<h2>Answer:</h2>

He is right that the energy of vaporization of 47 g of water s 106222 j.

<h3>Explanation:</h3>

Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.

In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.

It means for vaporizing 18 g, 40.65 kJ energy is needed.

So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ

Hence the student is right about the energy of vaporization of 47 g of water.

7 0
3 years ago
8.) If 396 g of Carbon Dioxide (CO2) are produced, what mass of Oxygen<br> (02) reacted? *
Brut [27]

Answer:

264g

Explanation:

C + O2 -> CO2

_g + _g -> 396g

396÷3=132

C (132g) + O2 (264g) -> CO2 (396g)

7 0
2 years ago
In the following reaction, if you wanted to produce more hydrochloric acid (HCl), what should you do? (2 points)
Paul [167]

Answer:

Add more H2O

Explanation:

Took the test

3 0
2 years ago
A metallurgical firm wishes to dispose of 1200 gallons of waste sulfuric acid whose molarity is 1.05 M. Before disposal, it will
Reil [10]

Answer:

The balanced chemical equation for this process:

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

$194.51 is the cost that the firm will incur from this use of slaked lime.

Explanation:

The balanced chemical equation for this process

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

Moles of sulfuric acid = n

Volume of sulfuric acid disposed = V = 1200 gallons = 3.785 × 1200 L = 4,542 L

1 gallon = 3.785 Liter

Morality of the sulfuric acid = M = 1.05 m

Molarity(M)=\frac{n}{V(L)}

n=m\times V=1.05 M\times 4,542 L=4,769.1 mol

According to reaction, 1 mol of sulfuric acid reacts with 1 mole of calcium hydroxide.Then 4,769.1 moles of sulfuric acid will recat with ;

\frac{1}{1}\times 4,769.1 mol=4,769.1 mol of calcium hydroxide

Mass of 4,769.1 moles of calcium hydroxide:

4,769.1 mol  74 g/mol = 352,913.4 g

= \frac{352,913.4 }{453.6} pounds =778.03 pounds

(1 pound = 453.6 grams)

Cost of 1 pound of slaked lime  = $0.25

Cost of 778.03 pounds of slaked lime  = $0.25 × 778.03 = $194.51

$194.51 is the cost that the firm will incur from this use of slaked lime.

3 0
3 years ago
Calculate the concentration of all species in a 0.160 m solution of h2co3.
enot [183]

             H₂CO₃ ⇔       HCO₃⁻   +     H⁺

I            0.160               0                 0

C            -x                  +x               +x

E          0.160-x          +x                +x

Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]

4.3 x 10⁻⁷ = x² / (0.160-x)   (x is neglected in 0.160-x = 0.160)

x² = 6.88 x 10⁻⁸

x = 2.62 x 10⁻⁴

             HCO₃⁻    ⇔            CO₃⁻²    +   H⁺

I          2.62 x 10⁻⁴               0                2.62 x 10⁻⁴

C          -x                            +x              +x

E       2.62 x 10⁻⁴ - x           +x              2.62 x 10⁻⁴ + x

Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]

5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)

x = 5.6 x 10⁻¹¹

Thus,

[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M

[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M

[CO₃⁻²] = 5.6 x 10⁻¹¹ M

[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M

[OH⁻] = 3.8 x 10⁻¹¹

8 0
2 years ago
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