Please help!! I'll name you brainliest!!

P O R F A V O R

ale4655 [162]

Answer:

a. 45×10³ kg

b. 1.25 kg

c. 5443200 s

d. 2.69×10⁻⁴ m/s²

e. 2.57×10⁻⁶N

f. 7.48×10⁻³ m /s²

g. 2.45 Pa

h. 10 m/s

Explanation:

The SI units are: kg, m, s, N, K, A, Pa, J and cd

a. 1 g is the mass for 1 cm³. We convert the m³ to cm³

45 m³. 1×10⁶ cm³ / 1 m³ = 45×10⁶ cm³

By the way, 45×10⁶ cm³ = 45×10⁶ g

We convert the g to kg → 45×10⁶ g . 1 kg / 1000 g = 45×10³ kg

b. As 1 g = 1 cm³, we convert the cm³ to g and then, the g to kg

1250 cm³ = 1250 g → 1250 g . 1kg / 1000 g = 1.25 kg

c. 1 day has 24 hours; 1 hour has 60 minutes; 1 minute has 60 seconds

1 hour has 3600 s. Then 24 h . 3600 s / 1 h = 86400 s

86400 s/d. 63 d = 5443200 s

d. 1 min² = 3600 s²

97 cm / 3600 s² = 0.0269 cm/s² / 100 = 2.69×10⁻⁴ m/s²

e. 927 g.cm / min² / 3600 s² = 0.2575 g.cm/s² → dyn

We need to convert dyn to N

1 dyn = 10⁻⁵N → 0.2575 dyn . 10⁻⁵N / 1dyn = 2.57×10⁻⁶N

f. 1 m/s² = 12960 km/h²

12960 km/h² . 1 m/s² / 97 km/h² = 7.48×10⁻³ m /s²

g. 2500 g/cm² . 1kg / 1000 g = 2.5 kg/cm²

1 Pa = 1.02kg/cm²

2.5 kg/cm² . 1 Pa / 1.02 kg/cm² = 2.45 Pa

h. 1 h = 3600 s

36 km / 3600 s = 0.01 km /s → 0.01 km . 1000 m / 1 km = 10

= 10 m/s

5 0

3 years ago

Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo

lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, $n_i=4$

Final orbit of electrons, $n_f=2$

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

$E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

Putting all the values we get :

$E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV$

We know that : $1\ eV=1.6\times 10^{-19}\ J$

So,

$E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J$

Energy of wave in terms of frequency is given by :

$E=hf$

$f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz$

Also, $c=f\lambda$

$\lambda$ is wavelength

So,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm$

Hence, this is the required solution.

4 0

3 years ago

How many moles of sodium hydrogen carbonate are present in 2.61 grams of this compound?

Sloan [31]

Molar mass of NaHCO_3

23+1+12+3(16)
36+48
84g/mol

Now

$\\ \sf\longrightarrow No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longrightarrow No\:of\; moles=\dfrac{2.61}{84}$

$\\ \sf\longrightarrow No\:of\:moles=0.0312mol$

7 0

2 years ago

When 27.50 liters of oxygen gas reacts with excess hydrogen gas at STP how many grams of water is produced?

telo118 [61]

<u>Answer: </u>The mass of water produced is 44.28 g

<u>Explanation:</u>

We are given:

Volume of oxygen gas = 27.50 L

At STP conditions:

22.4 L of volume is occupied by 1 mole of a substance

27.50 L of oxygen gas will be occupied by = $\frac{1mol}{22.4L}\times 27.50L=1.23mol$

The chemical equation for the formation of water follows:

$2H_2+O_2\rightarrow 2H_2O$

By the stoichiometry of the reaction:

If 1 mole of oxygen gas produces 2 moles of water

So, 1.23 moles of oxygen gas will produce = $\frac{2}{1}\times 1.23=2.46mol$ of water

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Molar mass of water = 18 g/mol

Plugging values in equation 1:

$\text{Mass of water}=(2.46mol\times 18g/mol)=44.28g$

Hence, the mass of water produced is 44.28 g

5 0

3 years ago

Someone please help will mark as brainliest

Sveta_85 [38]

Answer:

add me!!,

Explanation:

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7 0

3 years ago