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Natasha_Volkova [10]
3 years ago
10

How are the shape of sand dunes determined?using one word

Chemistry
1 answer:
SVEN [57.7K]3 years ago
8 0
"windweathered" because the use of wind weathering would bring the sand all around and it would be wind weathered
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Answer: In this compound, phosphorous and oxygen act together as one charged particle, which is connected to magnesium, the other charged particle.

Explanation:

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How is the relative abundance calculated?
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Atomic weights are weighted averages calculated by multiplying the relative abundance of each isotope by its atomic mass and then summing up all the products.
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NASA communicates with the Space Shuttle and International Space Station using Ku-band microwave radio. Suppose NASA transmits a
Tanya [424]

Answer:

A = 0.023 m

Explanation:

The relation between the frequency of a radiation and its wavelength is given by the following expression.

where,

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A = c/v = (3.00 x 10° m/s)/(13 x 10° s-") = 0.023

8 0
3 years ago
You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli
maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

M_1V_1=M_2V_2

where, M_1 is initial molarity and  M_2 is the molarity after dilution. Similarly,  V_1 is the volume before dilution and  V_2 is the volume after dilution.

Let's plug in the values in the equation:

1.25M(363mL)=0.50M(V_2)

V_2=\frac{1.25M(363mL)}{0.50M}

V_2=907.5mL

Volume of water added = 907.5mL - 363mL  = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0
3 years ago
How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric aci
horsena [70]

Answer:

22.7 g of CaCl₂ are produced in the reaction

Explanation:

This is the reaction:

CaCO₃ +  2HCl  →  CaCl₂  + CO₂  +  H₂O

Now, let's determine the limiting reactant.

Let's divide the mass between the molar mass, to find out moles of each reactant.

29 g / 100.08 g/m = 0.289 of carbonate

15 g / 36.45 g/m = 0.411 of acid

1 mol of carbonate must react with 2 moles of acid

0.289 moles of carbonate will react with the double of moles (0.578)

I only have 0.411 of HCl, so the acid is the limiting reactant.

Ratio is 2:1, so I will produce the half of moles, of salt.

0.411 / 2 = 0.205 moles of CaCl₂

Mol . molar mass = mass → 0.205 m . 110.98 g/m = 22.7 g

5 0
4 years ago
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