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3 years ago

9

Jill graphs the speed of a model train that is travelling at a speed of 15 kilometers per hour. Which graph shows the speed of t

he model train?

1 answer:

Tasya [4]3 years ago

8 0

Answer:c

Explanation:

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26. Which waves have wavelengths longer than those of visible light? Give an example of how each kind

Kipish [7]

Answer:

Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. Infrared waves have wavelengths longer than visible and shorter than microwaves, and have frequencies which are lower than visible and higher than microwaves.

Explanation:

8 0

3 years ago

Identifying polar regions within molecules is a very useful skill for chemists that helps us to think about the strength of inte

Ilia_Sergeevich [38]

bond in the acetic acid molecule is the most polar <span>c. c=o bond
</span>which bond in the methyl amine molecule is the most polar <span>c. n–h bond</span>

4 0

3 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]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

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

Convert 1000mg=__g 1L=__mL 160cm=__mm 1.4km=__m 109g=__kg 250m=__km 80cm=__m 75mL=__L 5.6m=__cm 6.5g=__mg 170.4m=__cm 564Dg=__g

Snowcat [4.5K]

Answer:

1000mg= 1g

1L= 1000 mL

160cm = 1600mm

1.4km= 1400m

109 g = 0.109kg

250m= 0.250 km

80cm= 0.8 m

75mL= 0.075L

5.6m= 560 cm

6.5g= 6500mg

170.4m= 17040 cm

564 Dg = 5640 g

58 dg = 5800 mg

600 L= 0.6 KL

0.0923Km= 92300 mm

Explanation:

1 mg = 1x10⁻³ g

1 g = 1000 mg

1 g = 10 dg

1 g = 1x10⁻³ kg

1 Dg = 10 g

1 dg = 100 mg

1 L = 1000 mL

1 L = 1x10⁻³ KL

1 mL = 1x10⁻³ L

1 km = 1000 m

1 km = 1x10⁶ mm

1 m = 1x10⁻³ km

1 cm = 1x10⁻² m

1 cm = 10 mm

5 0

3 years ago

Im posting seperate questions so just answer this one

vredina [299]

On the lab the text is kind of too far zoomed out so u can’t really read it it’s like blurry

6 0

2 years ago