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3 years ago

The inside window pane in your house feels very cold to touch on a winter night. How does the heat transfer?

Chemistry

1 answer:

kondaur [170]3 years ago

4 0

Answer:

its B

Explanation:

Because its can only go cold to hot. -hope this help :)

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NiSe2 + Ca3(PO4)2 = Ni3(PO4)2 + CaSe balance please

Aliun [14]

3Ni + Sn3(PO4)2 → Ni3(PO4)2 + 3Sn I think. Not for sure though

8 0

2 years ago

key A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration o

lorasvet [3.4K]

Answer:

2.41065 grams

Explanation:

Here we have to apply molarity, particularly in reference to the equation molarity = moles of solute / volume. I would like to rewrite this formula, but with respect to the units - grams = moles / Liters,

We can use molarity to determine the number of moles. After doing so, we can determine the mass of the solute with respect to the formula moles = mass / molar mass. The molar mass of NaCl is 58.44 grams.

_______________________________________________________

275 mL = 0.275 L,

Number of Moles of NaCl = 0.150 * 0.275 = 0.04125 moles,

Mass = 0.04125 * 58.44 = 2.41065 grams,

Solution - Mass of NaCl = 2.41065 grams

Hope that helps!

5 0

3 years ago

Read 2 more answers

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd

kenny6666 [7]

Answer:

The answer is " $= 0.078 \ kg \ H_2$ ".

Explanation:

calculating the moles in $CH_4 =\frac{PV}{RT}$

$=\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol$

Eqution:

$CH_4 +H_2O \to 3H_2+ CO \ (g)$

Calculating the amount of $H_2$ produced:

$= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2$

So, the amount of dihydrogen produced = $0.078 \frac{kg}{s}$

5 0

3 years ago

What is the major organic product obtained from the following reaction?

trapecia [35]

The correct option is (b)

NaNH2 is an effective base. It can be a good nucleophile in the few situations where its strong basicity does not have negative side effects. It is employed in elimination reactions as well as the deprotonation of weak acids.Alkynes, alcohols, and a variety of other functional groups with acidic protons, such as esters and ketones, will all be deprotonated by NaNH2, a powerful base.Alkynes are deprotonated with NaNH2 to produce what are known as "acetylide" ions. These ions are powerful nucleophiles that can react with alkyl halides to create carbon-carbon bonds and add to carbonyls in an addition reaction.Acid/base and nucleophilic substitution are the two types of reactions.Using the right base, terminal alkynes can be deprotonated to produce a carbanion.A good C is the acetylide carbanion.The acetylide carbanion can undergo nucleophilic substitution reactions because it is a potent C nucleophile. (often SN2) with 1 or 2 alkyl halides with electrophilic C to create an internal alkyne (Cl, Br, or I).Elimination is more likely to occur with 3-alkyl halides.It is possible to swap either one or both of the terminal H atoms in ethylene (acetylene) to create monosubstituted (R-C-C-H) and symmetrical (R = R') or unsymmetrical (R not equal to R') disubstituted alkynes (R-C-C-R').

Learn more about NANH2 here :-

brainly.com/question/12601787

#SPJ4

7 0

2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.