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3 years ago

What variables are used to calculate speed

Chemistry

2 answers:

Oksanka [162]3 years ago

6 0

S /\ Speed

mph, km/h, ft/s

Alja [10]3 years ago

5 0

Answer:

Calculating speed, distance and time

We can use formulas to model real-life situations.

For example, the formula for calculating speed is speed = distance ÷ time.

It is possible to calculate the speed, distance or time if you have the other two values.

Explanation:

You might be interested in

What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

REY [17]

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

5 0

2 years ago

If the relative rate of diffusion of ozone as compared to chlorine is 6:3 and further if the density of chlorine is 36 Find out

Alexandra [31]

Answer:

The density of ozone is 4.24.

Explanation:

The relation between the relative rate of diffusion and density is given by :

$r\propto \dfrac{1}{\sqrt d}$

The given ratio of the relative rate of diffusion of ozone as compared to chlorine is 6:3.

Let the density of ozone is d₂.

$\dfrac{r_1}{r_2}=\sqrt{\dfrac{d_2}{d_1}} \\\\\dfrac{6}{3}=\sqrt{\dfrac{d_2}{36}} \\\\3=\dfrac{\sqrt{d_2}}{6}\\\\d_2=\sqrt{18} \\\\d_2=4.24$

So, the density of ozone is 4.24.

6 0

3 years ago

The equation to calculate mean rate of reaction g/s

tatyana61 [14]

amount of product formed or amount of reactants used / time

8 0

3 years ago

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

2 years ago

A nuclear waste site. cesium-137 is a particularly dangerous by-product of nuclear reactors. it has a half-life of 30 years. it

hram777 [196]

The mass decay rate is of the form
$m(t) = m_{0} e^{-kt}$
where
m₀ = 3000 g,the initial mass
k = the decay constant
t = time, years.

Because the half-life is 30 years, therefore
$e^{-30k} = \frac{1}{2} \\\ -30k = ln(0.5) \\ k = \frac{ln(0.5)}{-30} =0.0231$

After 60 years, the mass remaining is
$m = 3000 e^{-0.0231*60} = 750 \, g$

Answer: 750 g

4 0

3 years ago