1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. 
0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4

6 0

3 years ago

Read 2 more answers

Someone please walk me through this. I googled way too much this semester and I'm paying for it on this test. Just tell me how t

alex41 [277]

Answer:

Explanation:

co(so4)4

4 0

3 years ago

Read 2 more answers

4NH3+5O2-4NO+6H2O how many moles of NH3 must react to produce 5.0 moles of NO?

tresset_1 [31]

Answer: 5.0 moles

Explanation:

From the equation, we see that for every 4 moles of ammonia consumed, 4 moles of nitrogen monoxide are produced (we can reduce this to moles of ammonia consumed = moles of nitrogen monoxide produced).

This means that the answer is 5.0 mol

7 0

1 year ago

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('

polet [3.4K]

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$

Next, by using the density of platinum we compute the volume:

$V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3$

Which computed in terms of the edge length is:

$V=a^3$

Therefore, the edge length turns out:

$a=\sqrt[3]{180cm^3}\\ \\a=5.65cm$

Best regards.

6 0

3 years ago