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Hunter-Best [27]

3 years ago

12

Can someone help me with 10 and 11 plz. I will mark them Brainliest

1 answer:

hjlf3 years ago

5 0

Answer:

10. Because they want to, and what they do is none of your buisness.

11. Firefighter-chemicals in fire-stopping have science.

Pharmacist-dealing with drugs and science was used to make them.

Chef-science involved in making food.-

Explanation:

Bruhh these were so easy do it yourself next time, it took me less than a minutte to do.

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On the diagram above, trace the path of a light ray through these materials

makvit [3.9K]

We need the diagram to answer the question

4 0

3 years ago

PLEASE ANSWER...........WILL MARK BRAINLIEST!!!!!!!

ANTONII [103]

Answer:

<h2>The Alkali metal halide may precipitate or there may be no change at all</h2>

Explanation:

Alkali metal cations are positively charged. Halogen anions are negatively charged. When a solution of Alkali metal cations is added to a solution of Halogen anions, there are two possibilities :

The alkali metal halide( salt formed from reation of the two ions) may precipitate if the Ionic product is higher than the Solubility product.
However, if it can remain in the solution, it will remain so. No chemical changes happen with respect to these both ions. Nothing willl happen.

There is no reaction happening in either of the cases because both species are already in ionic form before addition, hence they continue to be in this form.

3 0

3 years ago

What is the energy of a photon of infrared radiation with a frequency of 2.53 × 1012 Hz? Planck’s constant is

Novosadov [1.4K]

It is letter b or c nit d

6 0

1 year ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

3 years ago

1.What is the specific heat capacity of granite when 20 kg absorbs 237 000 J of heat energy, causing its temperature to increase

Natalka [10]

<u>Given</u> :

Amount = 20 kg
Heat energy absorbed = 237,000 J
Temperature change = 15 °C

<u>Formula applied</u> :

$\boxed {Q = mc \triangle T}$

Q = absorbed heat
m = mass
c = specific heat capacity
ΔT = temperature change

Let's solve for c !

⇒ 237,000 = 20 × c × 15

⇒ c = 237,000 ÷ 300

⇒ $\boxed {c = 790 J kg^{-1} K^{-1}}$

∴ The specific heat capacity of granite is <u>790 J kg⁻¹ K⁻¹</u>.

5 0

2 years ago