Ask question

Ask question

All categories

3 years ago

10

Which statement best describes the characteristics of white light?

1 answer:

netineya [11]3 years ago

6 0

Answer:

i believe its B

You might be interested in

If you have 3.54 g of H2, how many grams of NH3 can be produced?

morpeh [17]

3.54g H2 means you have 3.54 / 2 = 1.77 moles of H2, which means you have 1.77*2 = 3.54 moles of H atom.

Since 1 mole of NH3 has 3 moles of H, so you can produce 3.54 / 3 moles of NH3, which is (3.54/3) * (12+3) = 17.7 grams

5 0

3 years ago

A student uses a solution of potassium hydroxide to titrate a solution of nitric acid. Which question is the student trying to a

Mnenie [13.5K]

Answer:

A. What is the concentration of nitric acid?

Explanation:

its right

4 0

3 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

3 years ago

If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 C on a clorimetr liake hat in Fig

Dvinal [7]

<u>Answer:</u> The final temperature of water is 38.5°C

<u>Explanation:</u>

To calculate the amount of heat released or absorbed, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed = 1.506 kJ = 1506 J (Conversion factor: 1 kJ = 1000 J)

m = mass of water = 30 g

c = specific heat capacity of water = 4.184 Jl/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=T_2-26.5^oC$

Putting values in above equation, we get:

$1506J=30g\times 4.184J/g.^oC\times (T_2-26.5^oC)\\\\T_2=38.5^oC$

Hence, the final temperature of water is 38.5°C

4 0

4 years ago

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the or

Aloiza [94]

Answer:

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life Carbon 14 = 5730 year

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ hour^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}$

$\frac {[A_t]}{[A_0]}=0.007015$

<u>In percentage, the sample of C-4 remains = 0.7015 %</u>

3 0

3 years ago