What type of chemical reaction does this equation represent?<br> H3PO3

How is Earth's surface most likely to change when a river flows across a steep landscape over time?

ANTONII [103]

Answer:c

Explanation:

A valley will be carved out due to erosion.

4 0

2 years ago

Read 2 more answers

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)Â·2H2O. 2Na3(CO3)(HC

Leviafan [203]

Answer:

a. 92.4%

Explanation:

Based on the reaction:

2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)

To obtain the percent yield you need to obtain moles of trona and calculate thoeretical moles of Na₂CO₃, and the ratio of obtained moles / theoretical moles of Na₂CO₃ give percent yield, thus:

Moles of trona:

1.00 metric ton × (1x10³kg / 1 metric ton) × ( 1000moles /226.03 kg) = <em>4424 moles</em>

The theoretical moles of Na₂CO₃ that produce 4424 moles of trona are (Based on the reaction, 2 moles of trona produce 3 moles of Na₂CO₃):

4424 moles trona × (3 moles Na₂CO₃ / 2 moles trona) = <em>6636 moles of Na₂CO₃.</em>

The obtained moles of Na₂CO₃:

0.650 metric ton × (1x10³kg / 1 metric ton) × (1000 moles / 105.99kg) = <em>6133 moles</em>

The ratio of obtained moles / theoretical moles gives:

6133 moles / 6636 moles = 0.924 = <em>92.4%</em>

I hope it helps!

8 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$