Electric charge<br> A. discovery b. Radiation c.electricity d. Kinetic energy

In the Caribbean countries, general election is usually held every _ years

emmainna [20.7K]

it is held every 4-5 years

7 0

2 years ago

14 billion years (roughly the age of the universe) to seconds. (Assume there are 365 days in a year.)

fenix001 [56]

Answer:

4.41 × 10¹⁷ Seconds

Explanation:

For this problem we will first find out the number of days in 14 billion years.

Remember that 1 billion is equal to 1,000,000,000 or 10⁹. So, 14 billion years will be written as 14,000,000,000 or 1.4 × 10¹⁰ years.

Therefore, In

1 year there are = 365 years

So, In

1.4 × 10¹⁰ years there will be = X years,

Solving for X,

X = 1.4 × 10¹⁰ years × 365 days / 1 year

X = 5.11 × 10¹² days

Secondly, as there are 24 hours per day and each hour has 60 minutes and each minute has 60 seconds, therefore, the number of seconds per day are calculated as,

Number of Sec/day = 24 × 60 × 60

Number of Sec/day = 86400 sec/day

Hence, If,

1 day has = 86400 seconds

then,

5.11 × 10¹² days will contain = X seconds

Solving for X,

X = 5.11 × 10¹² days × 86400 sec / 1 day

X = 4.41 × 10¹⁷ Seconds

6 0

3 years ago

How many moles of na contain 7.88x1021 atoms of na

labwork [276]

Answer:

the answer to the qustion is 0.013089701 na

Explanation:

n/a

7 0

1 year ago

So guess what I did!!

Lapatulllka [165]

Answer:

what

Explanation:

what did u do

7 0

3 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

Electric charge A. discovery b. Radiation c.electricity d. Kinetic energy