What are elements that bond to form binary covalent compounds?<br> a) nonmetals<br> b) metals

Is a cube of salt being crushed before being stirred into water a physical or chemical change

Olin [163]

By crushing the salt, you are performing a physical change because you aren't altering the chemical makeup of the salt, just the physical form. Hope this helps! :)

6 0

3 years ago

Read 2 more answers

Pls answer this ASAP thank you

vaieri [72.5K]

Answer:

The anwer is not D the anwer is A

Explanation:

7 0

3 years ago

Hydrogen cyanide is used to prepare sodium cyanide, which is used in part to obtain gold from gold-containing rock. If a reac- t

Sliva [168]

Answer:

Mass of HCN produced = 6.75 g

Explanation:

Reaction is as follows:

$2NH_3+3O_2+2CH_4 \rightarrow 2HCN + 6H_2O$

First calculate the no. of moles of each chemical species.

molecular mass of $NH_3$ is 17 g/mol

No. of mol of $NH_3$ = 11.5/17 = 0.676 mol

Molecular mass of $O_2$ = 32 g/mol

No. of mol of $O_2$ = 12/32 = 0.375

Molecular mass of $CH_4$ = 16 g/mol

No. of mol of $CH_4$ = 10.5/16 = 0.656 mol

from the balanced chemical reaction, it is clear that 2-moles ammonia reacts with 3 moles oxygen and 2 moles methane to form 2 moles of HCN.

or, 1-mol ammonia reacts with 1.5 mol oxygen and 1 mol methane to form 1 mol of HCN.

Thus, no. of oxygen present is less than required and so it will act as limiting reagent.

From the chemical equation,

3 moles oxygen produces 2 moles HCN

or one mole oxygen produces (2/3) moles HCN

0.375 moles oxygen produces (2/3) × 0.375 HCN = 0.25 moles of HCN

molar mass of HCN = 27 g/mole

Mass = mol × molar mass

mass of HCN = 27 × 0.25 = 6.75 g

8 0

3 years ago

For each of the following unbalanced chemical equations suppose that exactly 50.0 g of each reactant is taken. Determine which r

Helen [10]

Answer:

1) Br2 is the limiting reactant.

Mass NaBr produced = 64.4 grams

2) CuSO4 is the limiting reactant

Mass Cu = 19.89 grams

Mass ZnSO4 = 50.54 grams

3) NH4Cl is the limiting reactant

Mass NaCl = 54.6 grams

Mass NH3 =15.9 grams

Mass H2O =16.8 grams

4) Fe2O3 is the limiting reactant

Mass Fe = 35.0 grams

Mass CO2 = 41.3 grams

Explanation:

1) Na+br2 ------------->Nabr

Step 1: Data given

Mass Na = 50.0 grams

Mass Br2 = 50.0 grams

Molar mass Na = 22.99 g/mol

Molar mass Br2 = 159.81 g/mol

Step 2: The balanced equation

2Na + Br2 → 2NaBr

Step 3: Calculate moles

Moles = mass / molar mass

Moles Na = 50.0 grams / 22.99 g/mol = 2.17 moles

Moles Br2 = 50.0 grams / 159.81 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

Br2 is the limiting reactant. It will completely be consumed (0.313 moles).

Na is in excess. There will react 2*0.313 = 0.626 moles

There will remain 2.17 - 0.626 = 1.544 moles

Step 5: Calculate moles NaBr

For 1 mol Br2 we'll have 2 moles NaBr

For 0.313 moles we'll have 0.626 moles NaBr

Step 6: Calculate mass NaBr

Mass NaBr = 0.626 moles * 102.89 g/mol

Mass NaBr = 64.4 grams

2) Zn+cuso4 -------------->Znso4+Cu

Step 1: Data given

Mass Zn = 50.0 grams

Mass CuSO4 = 50.0 grams

Molar mass Zn = 65.38 g/mol

Molar mass CuSO4 = 159.61 g/mol

Step 2: The balanced equation

Zn + CuSO4 → Cu + ZnSO4

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol = 0.765 moles

Moles CuSO4 = 50.0 grams / 159.61 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

CuSO4 is the limiting reactant. It will completely be consumed (0.313 moles).

Zn is in excess. There will react 0.313 moles

There will remain 0.765 - 0.313 = 0.452 moles

Step 5: Calculate moles products

For 1 mol Zn we need 1 mol CuSO4 to produce 1 mol Cu and 1 mol ZnSO4

For 0.313 moles CuSO4 we'll have 0.313 moles Cu and 0.313 moles ZnSO4

Step 6: Calculate mass products

Mass Cu = 0.313 moles * 63.546 g/mol = 19.89 grams

Mass ZnSO4 = 0.313 moles * 161.47 g/mol = 50.54 grams

3) NH4cl+NaOH -------------->NH3+H2O+NaCl

Step 1: Data given

Mass NH4Cl = 50.0 grams

Mass NaOH = 50.0 grams

Molar mass NH4Cl = 53.49 g/mol

Molar mass NaOH = 40.0 g/mol

Step 2: The balanced equation

NH4Cl + NaOH → NaCl + NH3 + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles NH4Cl = 50.0 grams / 53.49 g/mol = 0.935 moles

Moles NaOH = 50.0 grams / 40.0 g/mol = 1.25 moles

Step 4: Calculate limiting reactant

NH4Cl is the limiting reactant. It will completely be consumed (0.935 moles).

NaOH is in excess. There will react 0.935 moles

There will remain 1.25 - 0.935 = 0.315 moles

Step 5: Calculate moles products

For 1 mol NH4Cl we need 1 mol NaOH to produce 1 mol NaCl, 1 mol NH3 and 1 mol H2O

For 0.935 moles NH4Cl we'll have 0.935 moles NaCl, 0.935 moles NH3 and 0.935 moles H2O

Step 6: Calculate mass products

Mass NaCl = 0.935 moles * 58.44 g/mol = 54.6 grams

Mass NH3 = 0.935 moles * 17.03 g/mol = 15.9 grams

Mass H2O = 0.935 moles * 18.02 g/mol = 16.8 grams

4) Fe2O3+CO ------------>Fe+CO2

Step 1: Data given

Mass Fe2O3 = 50.0 grams

Mass CO = 50.0 grams

Molar mass Fe2O3 = 159.69 g/mol

Molar mass CO = 28.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe2O3 = 50.0 grams / 159.69 g/mol = 0.313 moles

Moles CO = 50.0 grams / 28.01 g/mol = 1.785 moles

Step 4: Calculate limiting reactant

Fe2O3 is the limiting reactant. It will completely be consumed (0.313 moles).

CO is in excess. There will react 3* 0.313 = 0.939 moles

There will remain 1.785 - 0.939 = 0.846 moles

Step 5: Calculate moles products

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe, 3 moles CO2

For 0.313 moles Fe2O3 we'll have 0.626 moles Fe and 0.939 moles CO2

Step 6: Calculate mass products

Mass Fe = 0.626 moles * 55.845 g/mol = 35.0 grams

Mass CO2 = 0.939 moles * 44.01 g/mol = 41.3 grams

8 0

3 years ago

What makes the Arrhenius acid and an Arrhenius base definition incomplete or lacking

Kryger [21]

Answer:

See Explanation

Explanation:

The Arrhenius acid-base theory defines an acid as a compound which when added into water increases the hydronium ion (H₃O⁺) concentration and the base as a compound which when added into water increases the hydroxide (OH⁻) ion concentration. As such, an acid-base reaction is limited to proton transfer to only OH⁻ ions forming water. Such would imply that all acid-base reactions produce water only in addition to a salt. This is not always the case as conjugate base anions for many substances can receive proton transfer.

Example: The reaction HOAc + NaCN => HCN + OAc- will occur in aqueous media because the proton (H⁺) on acetic acid (HOAc) will transfer to the cyanate ion forming hydrocyanic acid (HCN). Such occurs because the CN⁻ ion is a stronger conjugate base than the acetate ion (OAc⁻) and forms the more stable weak acid. Such is the basis of the Bronsted-Lowry Acid-Base system and states that an acid (proton donor) will transfer its ionizable hydrogen to a conjugate base (proton acceptor) if the transfer forms a weaker acid.

7 0

3 years ago

What are elements that bond to form binary covalent compounds? a) nonmetals b) metals