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3 years ago

20 POINTSSS PLS HELP! Which of the following things threaten our fresh water supply? Check all that apply.

Chemistry

2 answers:

zubka84 [21]3 years ago

7 0

Answer: D and B

Hope this helps ;)

joja [24]3 years ago

3 0

Fertilizer use the run on can go into fresh water

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What is the volume of 45.6g of silver if the density of silver is 10.5g/mL? A. 4.34mL B. 479mL C. 0.23mL

nirvana33 [79]

Explanation:

first you get moles of silver

n=m/M

hence you add no of moles to this equation

c=nv

v=n/c

8 0

4 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

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

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago

What would the products of a double-replacement reaction between KBr and CaO be? (Remember: In double-replacement reactions, the

11111nata11111 [884]

ANSWER

$\begin{gathered} \text{ 2KBr }+\text{ CaO }\rightarrow\text{ K}_2O\text{ }+\text{ CaBr}_2 \\ Option\text{ B} \end{gathered}$

EXPLANATION

Given that;

The two reactants are KBr and CaO

Double replacement reaction is a type of chemical reaction that occur when two reactants exchange cations and anions to yield new products.

$\text{ 2KBr + CaO }\rightarrow\text{ K}_2O\text{ + CaBr}_2$

Therefore, the resulting products of the given data are K2O + CaBr2

The correct answer is option B

4 0

1 year ago

Can someone help me

iogann1982 [59]

Answer:

i cant see it or i would

Explanation:

5 0

3 years ago

The density of an aqueous solution of nitric acid is 1.64 g/mL and the concentration is 1.85 M. What is the concentration of thi

galina1969 [7]

Answer:

Mass % of the solution = 7.1067 %

Explanation:

Given :

Molarity of nitric acid solution = 1.85 M

Density of the solution = 1.64 g/mL

Molarity of a solution is defined as the number of moles of solute present in 1 liter of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Lets, consider the volume of the solution = 1 L

Thus,

Moles of nitric acid present in the solution:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles of Nitric acid=Molarity \times {Volume\ of\ the\ solution}$

So,

Moles of Nitric acid = 1.85 moles

Molar mass of nitric acid = 63 g/mol

The mass of Nitric acid can be find out by using mole formula as:

$moles=\frac{Mass\ taken}{Molar\ mass}$

Thus,

$Mass\ of\ Nitric\ acid=Moles \times Molar mass}$

$Mass\ of\ Nitric\ acid=1.85 g \times 63 g/mol}$

Mass of Nitric acid = 116.55 g

Also,

$Density=\frac{Mass}{Volume}$

Given : Density = 1.64 g/mL

Also, 1 L = 10³ mL

Volume of the solution is 1000 mL

So, mass of the solution:

$Mass\ of\ the\ solution=Density \times {Volume\ of\ the\ solution}$

$Mass\ of\ the\ solution=1.64 g/mL \times {1000 mL}$

Mass of the solution = 1640 g

Mass % is defined as the mass of solute in 100 g of the solution. The formula for the calculation of mass % is shown below:

$Mass \% =\frac{Mass\ of\ the\ solute}{Mass\ of\ the\ solution} \times {100}$

So,

$Mass \%=\frac{116.55}{1640} \times {100}$

Mass % = 7.1067 %

6 0

3 years ago