What is the compound name for Bel?

What is a property of bases?

ycow [4]

Answer:

hi

Explanation:

the answer is slippery touch

I hope this helps

please give brainest

8 0

3 years ago

What mass of aluminum (in g) would be required to completely react with 1.00 L of 0.450 M HBr in the following chemical reaction

Mama L [17]

Answer:

4.0473g. Attached in image are the calculations I used to reach this answer.

3 0

2 years ago

Arrange the steps of the scientific method in the proper order.

JulsSmile [24]

Answer:

Identify the Problem

Observe and Record

Research the Problem

Make a Hypothesis

Test the Hypothesis

Arrive at a Conclusion

Hope this Helps!!

3 0

3 years ago

Read 2 more answers

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

ss7ja [257]

Answer:

1.89 of Sodium Carbonate

3.94 g of Silver Carbonate

2.43 g of Sodium Nitrate

Zero grams of Silver Nitrate

Explanation:

We have to start with the reaction:

$AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~NaNO_3$

Now, we can balance the reaction:

$2AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~2NaNO_3$

Now, we have to calculate the limiting reagent and we have to follow a few steps:

1) Convert to moles (using the molar mass of each compound)

2) Divide by the coefficient of each reactive (given by the balanced reaction)

Convert to moles

$3.40~g~Na_2CO_3\frac{105.98~g~Na_2CO_3}{1~mol~Na_2CO_3}=0.032~mol~Na_2CO_3$

$4.86~g~AgNO_3\frac{169.8~g~AgNO_3}{1~mol~AgNO_3}=0.0286~mol~AgNO_3$

Divide by the coefficient

$\frac{0.032~mol~Na_2CO_3}{1}=0.032$

$\frac{0.0286~mol~AgNO_3}{2}=0.0143$

The smallest value is for $AgNO_3$ , therefore the 4.86 g of $AgNO_3$ .

Now we can calculate the amount of compounds produced is we follow a few steps:

1) Use the molar ratio

2) Convert to moles (using the molar mass of each compound)

Amount of Silver Carbonate

$0.0286~mol~AgNO_3\frac{1~mol~AgCO_3}{2~mol~AgNO_3}\frac{275.74~g~AgCO_3}{1~mol~AgCO_3}=3.94~g~AgCO_3$

Amount of Sodium Nitrate

$0.0286~mol~AgNO_3\frac{2~mol~NaNO_3}{2~mol~AgNO_3}\frac{84.99~g~NaNO_3}{1~mol~NaNO_3}=2.43~g~NaNO_3$

Amount of Sodium Carbonate (Excess reactive)

$0.0286~mol~AgNO_3\frac{1~mol~NaCO_3}{2~mol~AgNO_3}\frac{105.98~g~NaCO_3}{1~mol~NaCO_3}=1.51~g~NaCO_3$

$3.4~g~NaCO_3-1.51~g~NaCO_3=1.89~g~NaCO_3$

Amount of Silver Nitrate

All the silver nitrate would be consumed in the reaction

I hope it helps!

7 0

4 years ago

Please write the answer in format below

jeyben [28]

Answer:

0.95 L

Explanation:

Step 1: Given data

Concentration of the Mg(NO₃)₂ solution (C): 0.32 M (0.32 mol/L)

Mass of Mg(NO₃)₂ (solute): 45 g

Step 2: Calculate the moles corresponding to 45 g of Mg(NO₃)₂

The molar mass of Mg(NO₃)₂ is 148.33 g/mol.

45 g Mg(NO₃)₂ × 1 mol Mg(NO₃)₂ /148.33 g Mg(NO₃)₂ = 0.303 mol Mg(NO₃)₂

Step 3: Calculate the volume of solution that contains 0.303 moles of Mg(NO₃)₂

The concentration of the solution is 0.32 M, that is, there are 0.32 moles of Mg(NO₃)₂ per liter of solution.

0.303 mol Mg(NO₃)₂ × 1 L Solution / 0.32 mol Mg(NO₃)₂ = 0.95 L

4 0

3 years ago