Is distilled water a compound, element, or mixture?

Match the symbol for each element with the name of the element. 1. hydrogen K 2. helium Mg 3. sodium He 4. magnesium H 5. potass

insens350 [35]

Answer:

1. hydrogen - H

2. helium - He

3. sodium - Na

4. magnesium - Mg

5. potassium - K

Explanation:

Hydrogen is the element of group 1 and first period. The atomic number of hydrogen is 1 and the symbol of the element is H.

The electronic configuration of the element hydrogen is:-

$1s^1$

Helium is the element of group 18 and first period. The atomic number of helium is 2 and the symbol of the element is He.

The electronic configuration of the element helium is:-

$1s^2$

Sodium is the element of group 1 and third period. The atomic number of sodium is 11 and the symbol of the element is Na.

The electronic configuration of the element sodium is:-

$1s^22s^22p^63s^1$

Magnesium is the element of group 2 and third period. The atomic number of magnesium is 12 and the symbol of the element is Mg.

The electronic configuration of the element magnesium is:-

$1s^22s^22p^63s^2$

Potassium is the element of group 1 and forth period. The atomic number of potassium is 19 and the symbol of the element is K.

The electronic configuration of the element potassium is:-

$1s^22s^22p^63s^23p^64s^1$

5 0

3 years ago

What is the mass of 7 x 10^28 atoms of Fe?

Nostrana [21]

Answer:

6 x 10⁶ g Fe

Explanation:

Step 1: Set up dimensional analysis

7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)

Step 2: Multiply, divide, and cancel out units

atoms Fe and atoms Fe cancel out.

mol Fe and mol Fe cancel out.

We should be left with g Fe.

7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe

116279(55.85) = 6.49 x 10⁶ g Fe

Step 3: Sig figs

There is only 1 sig fig in this problem.

6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe

4 0

3 years ago

What is the mass in grams of 6.12 x 10^20 molecules of caffeine, C8H10N4O2?

baherus [9]

$6,02*10^{23} \ \ \ \ \ \ \rightarrow \ \ \ \ \ 194,2g\\
6,12 * 10^{20} \ \ \ \ \ \ \rightarrow \ \ \ \ \ x\\\\
x=\frac{6,12*10^{20}*194,2g}{6,02*10^{23}}\approx197*10^{-3}g=0,197g$

6 0

3 years ago

Use the following information to answer the following question.

Irina-Kira [14]

Answer:

For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%

Cu + 2 AgNOâ‚ â†’ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver

12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:

38.1-g/43.2-g x 100% = 88.2%

Explanation:

5 0

3 years ago

For the reaction: Cr2O7^2- + NH4^+ → Cr2O3 + N2, which is the oxidizing agent?

lord [1]

Answer:

(NH4)2Cr2O7

Explanation:

Hope this somehow helped.

3 0

3 years ago