Which landform is located at the mouth of the river?

1 answer:

3 0

Landforms. At the mouth of a river, the change in flow condition can cause the river to drop any sediment it is carrying. This sediment deposition can generate a variety of landforms, such as deltas, sand bars, spits, and tie channels.

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How many moles are in 7.1x10^21 atoms of iron?

Nitella [24]

Answer:

0.011 moles

Explanation:

There are about 6.02*10^23 atoms in a mole, so in the given sample, there are

$\frac{7.01 \times {10}^{21} }{6.02 \times {10}^{23} }$

which is about 0.011 moles.

8 0

2 years ago

10. How many g of Cu(OH)2 can be made from 9.1 x 1025 atoms of O?

il63 [147K]

Molar mass Cu(OH)₂ = 97.561 g/mol

97.561 g Cu(OH)₂ --------------- 6.02x10²³ atoms
? g Cu(OH)₂ -------------------- 9.1x10²⁵ atoms

mass = 9.1x10²⁵ * 97.561 / 6.02x10²³

mass = 8.87x10²⁷ / 6.02x10²³

mass = 14734.2 g

hope this helps!

8 0

3 years ago

Why is it necessary to use clean sponge and cleaning rags in dishwashing?

NARA [144]

Answer:

cloths/sponges can harbor harmful pathogens and spread germs if not cleaned frequently. A damp, smelly dish cloth/sponge is telling you germs are multiplying!

Explanation:

hope this is able to help you :)

6 0

2 years ago

Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container

Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

Explanation:

Here Pressure in a container is given as

$P=\frac{1}{3} \rho$

Here

P is the pressure which is to be calculated

ρ is the density of the gas which is to be calculated as below

$\rho =\frac{mass}{Volume of container}$

Here

mass is to be calculated for 75 gas phase molecules as

$m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g$

Volume of container is 0.5 lts

So density is given as

$\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3$