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2 years ago

Why can't astronomers take a picture of a black hole?

Chemistry

2 answers:

Yakvenalex [24]2 years ago

7 0

Answer:

Explanation:

san4es73 [151]2 years ago

5 0

Answer:

Gravity is so strong that even light cannot escape a black hole.

Explanation:

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The specific heat capacity of a pure substance can be found by dividing the heat needed to change the temperature of a sample of

kupik [55]

Answer:

See below

Explanation:

ΔQ = m c T ΔQ = heat required(J) m = mass (g) T = C° temp change

c = heat capacity in J/g-C

4 0

2 years ago

Select all the correct answers

Alborosie

Answer:

Explanation:

4 0

2 years ago

A solution is prepared by dissolving 4.40 g of KSCN in enough water to make 200 mL of solution. What is the concentration of the

attashe74 [19]

First find the number of moles of KSCN is in 4.40g of KSCN 1. Number of moles = mass given (4.40g) / molecular weight(MW) (KSCN) - to calculate MW, look for each element in the periodic table and add their atomic weight 2. Convert 200mL to liters 3. Concentration = number of moles/ volume in liters from (#2)

4 0

2 years ago

Read 2 more answers

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

3 years ago

Consider this mechanism:

DerKrebs [107]

Answer:

$ClO^- + I^-\rightarrow IO^- + Cl^-$

Explanation:

Given a reaction mechanism, we will typically have catalysts and intermediates which will not be observed in the final balanced overall reaction. In order to obtain a net reaction, we need to add all the separate steps of the given reaction.

First of all, the reactants are combined followed by a combination of products. The repeating species on both sides of the final equation are then canceled out. Let's sum everything we have in our steps:

$ClO^- + H_2O + I^- + HClO + OH^- + HIO\rightarrow HClO + OH^- + HIO + Cl^- + H_2O + IO^-$

Repeating species that can be canceled out are:

$H_2O,~HClO,~OH^-,~HIO$

This leaves a net reaction of:

$ClO^- + I^-\rightarrow IO^- + Cl^-$

5 0

2 years ago