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2 years ago

What is the only nonmetal with four valence electrons.

Chemistry

1 answer:

Alja [10]2 years ago

3 0

Carbon is a non metal and also has four valence electrons.

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If you had 100 ml of juice, how many milliliter would de fruit juice?

Julli [10]

Since the measurement is not changing, the answer is 100 mL. Hope this helps.

8 0

3 years ago

How many oxygen atoms are in one molecule of baking soda

8_murik_8 [283]

NaHCO3 = No. of atoms are 1 sodium + 1 Hydrogen + 1 carbon + 3 oxygens = 6 atoms per molecule.

so a total of 3 oxygens

4 0

2 years ago

Explain the difference between qualitative and quantitative properties.

SVETLANKA909090 [29]

Qualitative properties are properties that are observed and can generally not be measured with a numerical result. They are contrasted to quantitative properties which have numerical characteristics.

3 0

2 years ago

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

Fiesta28 [93]

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

$N= N_0(\frac{1}{2})^n$

where

$n = \frac{t}{t_{\frac{1}{2}}}$

here

t = 458,000 years

$t_{\frac{1}{2}}$ = 229,000

$\frac{t}{t_{\frac{1}{2}}}$ = \ $\frac{ 458,000}{229,000}$

n = $\frac{t}{t_{\frac{1}{2}}}$ = 2.000

Now substituting the values

$1.675 = N_0(\frac{1}{2})^{2.000}}$

$1.675 = N_0\times (0.2500)$

$N_0= \frac{1.675}{0.2500}$

$N_0=6.70$

3 0

2 years ago

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

xenn [34]

Answer:

$5.225x10^{22}atoms\ I$

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

$n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2$

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

$atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I$

Best regards!

4 0

3 years ago