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3 years ago

What effect do calcium ions have on the water in ecosystems?

Chemistry

2 answers:

Marysya12 [62]3 years ago

6 0

It's because they add to its hardness . Calcium is naturally dissolved in water

BlackZzzverrR [31]3 years ago

4 0

Answer:

They add to its hardness.

Explanation:

Hello,

Water in ecosystems are likely to have dissolved metallic cations such those alkalies and alkaline earth metals form when present in aqueous systems, in this manner, more specifically speaking, calcium ions promote the formation of unsoluble salts such as calcium carbonate, which lead to the increase of the water's hardness that accounts for such cation's content into the water. Therefore, among the following options:

A. They reduce its pH.

B. They decrease its salinity.

C. They add to its hardness.

D. They increase its acidity.

The correct is C. They add to its hardness.

Best regards.

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A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = $\frac{18.0 g}{60 g/mol}=0.3 mol$

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

$Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}$

Molarity of the urea solution ;

$M=\frac{0.3 mol}{0.200 L}=1.50 M$

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

$m=d\times V=0.95 g/mL\times 200.0 mL=190 g$

$m = 190 g = 190 \times 0.001 kg = 0.19 kg$

(1 g = 0.001 kg)

Molality of the urea solution ;

$Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

$m=\frac{0.3 mol}{0.19 kg}=1.58 m$

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0

2 years ago

Is water wet? I give you a lot of points

UkoKoshka [18]

i mean technically, no. only because water is water and water makes things wet. you know? unless you pour water onto water then idk honestly, truly...

6 0

3 years ago

Read 2 more answers

PLS HELP-CHEMISTRY: What is the molar concentration of an ammonia solution in which there are 0.1557 grams

harina [27]

Answer: M = 0.036 M

Explanation: Solution attached:

First convert mass of NH3 to moles

Next convert volume in mL to L

Use the formula for Molarity

M = n / L

4 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.