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erma4kov [3.2K]
2 years ago
8

There are 24 hours in a day, 60 minutes in an hour, 7 days in a week, and 4 weeks in a month? using these conversions how many m

inutes are there in 1 month?
(i would appreciate it if someone gave a step by step explanation)
Chemistry
1 answer:
Serggg [28]2 years ago
5 0

Explanation:

From your information above, assume 1 month is 28 days

1 days=24hours ; 28 days=672 hours (just multiply 1 days with 28)

1 hour=60minutes ; 672 hours=40,320 minutes

.. 1 month=40,320 minutes

You might be interested in
Is the law of conservative mass observed in this equation CaCO3 + 2HCI -->CaCI2 +H2O + CO2
pychu [463]

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:

  • Ca ⇒ 1 on the left, 1 on the right.
  • C ⇒ 1 on the left, 1 on the right.
  • O ⇒ 3 on the left, 3 on the right.
  • H ⇒ 2 on the left, 2 on the right.
  • Cl ⇒ 2 on the left, 2 on the right.

As the numbers for all elements involved are the same, the law is observed in the given equation.

8 0
3 years ago
Why do you think different liquids have different freezing points
Ganezh [65]
I think that different liquids have different freezing points because every liquid consists of different atoms and different things that make up the atom causing them to have different freezing points.
5 0
3 years ago
1.5 moles are present in 60.0 grams of calcium.
Andrew [12]

Answer:

True.

Explanation:

To know which option is correct, let us calculate the number of mole present in 60g of calcium. This is illustrated below:

Mass of Ca = 60g

Molar Mass of Ca = 40g/mol

Number of mole Ca =....?

Number of mole = Mass/Molar Mass

Number of mole of Ca = 60/40

Number of mole Ca = 1.5 moles.

From the calculations made above, we can see that 1.5 moles are present in 60.0 grams of calcium

3 0
3 years ago
The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14
Elena-2011 [213]

Answer:

Age of the atifact is 4.2\times 10^{2} years

Explanation:

  • For first -order radioactive decay- A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}
  • A_{t} represents activity of radioactive nuclide after t time, A_{0} represents initial activity of radioactive nuclide and t_{\frac{1}{2}} represents half-life
  • Here, A_{t}=19Bq, A_{0}=20Bq and t_{\frac{1}{2}}=5.73\times 10^{3}years

Plug-in all the given values in the above equation-

19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}

or, t=4.2\times 10^{2}

So, age of the atifact is 4.2\times 10^{2} years

6 0
3 years ago
A 0.5376 g sample of an unknown compound is found to contain 0.3044 g of carbonate. Could this compound be calcium carbonate?
shepuryov [24]
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams

One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams

Based on the above calculations, the sample is not CaCO3
6 0
3 years ago
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