What astronomical equiptment did hypatia use?

What is the balanced equation for the combustion of magnesium?

Lostsunrise [7]

Answer:

2Mg + O2 → 2MgO

Explanation:

In all conbustion you should know, that reactans are an specific compound and O2, so the products must be CO2 and H2O, or in this case, the corresponding oxide.

6 0

3 years ago

Which element would make a poor, brittle conductor? use the periodic table for help. a. sulfur b. boron c. cobalt d. calcium?

Greeley [361]

I think the answer is C or B

8 0

2 years ago

(a)-Use Lewis symbol store present there action that occurs between Ca and F atoms. (b)-What is the chemical formula of the most

Charra [1.4K]

Answer:

(a) The Lewis structure is shown in the image below.

(b) $CaF_2$

(c) Calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond.

(d) Calcium atom loses electrons.

Explanation:

Calcium is the element of second group and forth period. The electronic configuration of Calcium is - 2, 8, 8, 2 or $1s^22s^22p^63s^23p^64s^2$

There are 2 valence electrons of Calcium.

Fluorine is the element of group 17 and second period. The electronic configuration of the element fluorine is - 2, 7 or $1s^22s^22p^5$

There is 1 valence electron of fluorine.

The Lewis structure is drawn in such a way that the octet of each atom is complete.

Thus, calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.

Thus, the formula of calcium chloride is $CaF_2$ .

(a) The Lewis structure is shown in the image below.

(b) $CaF_2$

(c) Calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond.

(d) Calcium atom loses electrons.

5 0

2 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

2 years ago

Read 2 more answers

What made alchemy unscientific in its practices?

anzhelika [568]

(B) because it couldn’t turn in to good

7 0

3 years ago

Read 2 more answers