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3 years ago

Hof metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the

Chemistry

1 answer:

Rus_ich [418]3 years ago

3 0

Answer:

The plate would only begin to slowly cool down

Explanation:

As the plate is introduced to room temperature the plate kind of stimulates oxidation in a way as in the plate is no longer in a state of continuous heat.

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How many grams are there in 7.4 x 1023 molecules of H2SO4?

bazaltina [42]

7.4x10^23 = molecules of silver nitrate sample

6.022x10^23 number of molecules per mole (Avogadro's number)

Divide molecules of AgNO3 by # of molecules per mol

7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)

(I leave off the x10^23 because they both will divide out)

Use your periodic table to find the molar weight of silver nitrate.

107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3

Now multiply your moles of AgNO3 with your molar weight of AgNO3

1.229mol x 169.868g/mol = 208.767g AgNO3

4 0

4 years ago

A sample that contains only SrCO3 and BaCO3 weighs 0.800 g. When it is dissolved in excess acid, 0.211 g car- bon dioxide is lib

MrRissso [65]

Answer:

53.9%

Explanation:

1 mole of BaCO₃ yields 1 mole of CO₂,

1 mole of SrCO₃ yields 1 mole of CO₂

m₁ = mass of BaCO₃

m₂ = mass SrCO₃

molar mass of SrCO₃ = 147.63 g/mol

molar mass of BaCO₃ = 197.34 g/mol

molar mass of CO₂ = 44.01 g/mol

mole of CO₂ in 0.211 g = 0.211 g / 44.01 = 0.00479

mole of BaCO₃ = m₁ / 197.34

mole of SrCO₃ = m₂ / 147.63

mole of BaCO₃ + mole of SrCO₃ = 0.00479

(m₁ / 197.34) + (m₂ / 147.63) = 0.00479

147.63 m₁ + 197.34 m₂ = 139.55

m₁ + m₂ = 0.8

m₁ = 0.8 - m₂

147.63 (0.8 - m₂) + 197.34 m₂ = 139.55

118.104 - 147.63 m₂ + 197.34 m₂ = 139.55

49.71 m₂ = 139.55 - 118.104 = 21.446

m₂ = 21.446 / 49.71 = 0.431

the percentage of m₂ ( SrCO₃ ) in the sample = 0.431 / 0.8 = 0.539 × 100 = 53.9%

5 0

3 years ago

How many Liters of 0.50M HCl are needed to neutralize 0.050L of 0.101M Ba(OH)2?

Aleksandr-060686 [28]

Answer:

$V_{HCl}=5.05x10^{-3}L$

Explanation:

Hello,

In this case, since hydrochloric acid and barium hydroxide are in a 2:1 molar ratio, for the neutralization, the following moles equality must be obeyed:

$2*n_{HCl}=n_{Ba(OH)_2}$

In such a way, in terms of molarities and volumes, we can compute the required volume of hydrochloric acid as shown below:

$2*M_{HCl}V_{HCl}=M_{Ba(OH)_2}V_{Ba(OH)_2}\\\\V_{HCl}=\frac{M_{Ba(OH)_2}V_{Ba(OH)_2}}{2M_{HCl}} =\frac{0.101M*0.050L}{2*0.50M} \\\\V_{HCl}=5.05x10^{-3}L$