This problem is providing us with the **mass **of a **sample **containing **osmium **and **oxygen **and the **mass **of **former**, so the **empirical formula** of this **substance** is **required **and found to be **OsO₄** after the following.

<h3>Empirical formulas</h3>

In chemistry, **chemicals **have **specific formulas **showing the **number **and **type **of **atoms present**, via the **elements'** **symbols **and **subscripts **in the **formula**. In this case, since this compound **contains** **2.16** g of **osmium**, one can **calculate **the **mass** of **oxygen **with:

In such a way, one can then **calculate **the **moles** of the **both** of the **elements**:

Then, we **divide **by the **fewest moles **to get their **subscripts **in the **formula**:

Which means the **oxide **has a **formula **of **OsO₄**.

Learn more about **empirical formulas**: brainly.com/question/11588623

1. True

2. False

3. False

4. True

5. True

6. False

7. True

8. True

9. True

10. Base

11. Acid

12. Base

13. Acid

14. Neutral

**Answer:**

Halogens always form anions, alkali metals and alkaline earth metals always form cations

**Explanation:**

**Answer: The partial pressure of **** is 0.35 atm.**

**Explanation:**

**Given:** Total pressure = 0.98 atm

Partial pressure of = 0.48 atm

Partial pressure of Ar = 0.15 atm

Partial pressure of = ?

**Total pressure is the sum of partial pressure of each component present in a mixture of gases.**

Hence, partial pressure of is calculated as follows.

Total pressure =

Substitute the values into above formula as follows.

Thus, we can conclude that **the partial pressure of **** is 0.35 atm.**

**Answer:**

The answer is "**208 mL**".

**Explanation:**

In dilutions of different concentration, its initial volume for glucose solution V1 can be obtained by using the following formula, i.e.

The solution of 520 mL final glucose solution thus requires 208 mL of glucose solution of .