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4 years ago

Which of the following statements about the sun's features are true?

Chemistry

2 answers:

hodyreva [135]4 years ago

7 0

Answer:

the answer is B Solar flares form when prominence's connect

Explanation:i took the quiz

Aleonysh [2.5K]4 years ago

4 0

Solar flares form when prominences connect I think

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How many grams of Chlorine are there in 1000 g of Iron(III) chloride?

Daniel [21]

Approx.
15
⋅
g
of metal.
Explanation:
Moles of iron oxide,
F
e
2
O
3

=

21.6
⋅
g
159.69
⋅
g
⋅
m
o
l
=

0.135
⋅
m
o
l
with respect to the oxide.
But by the composition of the oxide, there are thus
2
×
0.135
⋅
m
o
l
×
55.8
⋅
g
⋅
m
o
l
−
1

iron metal

=

?
?
g
.
An extraordinary percentage of our budgets goes into rust prevention;
iron(III) oxide
is only part of the redox chemistry. Once you put up a bridge or a skyscraper, which is inevitably steel-based structure, it will begin to corrode.

3 0

4 years ago

Read 2 more answers

What is the mole of 125 mg of Na? How many Na atoms?

Crank

(125 mg Na) x (1 g/1000mg) x (1 mol of Na/22.99 g) = 5.43 E -3 mols of Na

5.43 E -3 mols x 6.022 E 23 = 3.27 E 21 Na atoms

7 0

3 years ago

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq) ∆ HbCO(aq) + O2(aq) a. Use

gayaneshka [121]

The question is incomplete, complete question is;

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction:

$HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)$

Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.

$Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8$

$Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306$

Answer:

The equilibrium constant for the given reaction is 170.

Explanation:

$Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8$

$K_1=\frac{[HbO_2]}{[Hb][O_2]}$

$[Hb]=\frac{[HbO_2]}{[K_1][O_2]}$ ..[1]

$Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306$

$K_2=\frac{[HbCO]}{[Hb][CO]}$ ..[2]

$HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)$

$K_c=\frac{[HbCO][O_2]}{[HbO_2][CO]}$ ..[3]

Using [1] in [2]:

$K_2=\frac{[HbCO]}{\frac{[HbO_2]}{[K_1][O_2]}\times [CO]}$

$K_2=K_1\times \frac{[HbCO][O_2]}{[HbO_2][CO]}$

$K_2=K_1\times K_c$ ( using [3])

$306=1.8\times K_c$

$K_c=\frac{306}{1.8}=170$

The equilibrium constant for the given reaction is 170.

7 0

3 years ago

Elzmall.org Bookmark

soldier1979 [14.2K]

I believe the answer is PD 6

4 0

3 years ago

Radioactive carbon-14 has a half life of 5730 years. Suppose a peice of wood has a decay rate of 15 disintegrations per minute.

Tom [10]

Answer:

$\large \boxed{\text{2920 yr}}$

Explanation:

Two important formulas in radioactive decay are

$(1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt$

1. Calculate the decay constant k

$\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{1530 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{1530 yr}}\\\\& = & 4.530 \times 10^{-4} \text{ yr}^{-1}\\\end{array}$

2. Calculate the time

$\begin{array}{rcl}\ln \left(\dfrac{N_{0}}{N}\right) &= &kt \\\\\ln \left(\dfrac{15}{4}\right) &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\\\\ln 3.75 &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\t &= &\dfrac{\ln 3.75}{4.530 \times 10^{-4} \text{ yr}^{-1}}\\\\& = & \textbf{2920 yr}\\\end{array}\\\text{It would take $\large \boxed{\textbf{2920 yr}}$ for the rate to decrease to 4 dis/min.}$

5 0

4 years ago