Choose the correct statement for this question: If two objects have different temperatures, in which direction will heat move? For how long?

1. Heat moves from hot to cold objects/areas, and it never stops.
2. Heat moves from cold to hot object/areas.
3.Heat moves from hot objects/areas to cold areas and stops moving once it reaches a thermal equilibrium.

Answer:

chemical

Explanation:

Answer:

150.1 mL

Explanation:

Step 1: Given data

Step 2: Calculate the volume of the sample of benzene

Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.

ρ = m/V

V = m/ρ

V = 131.9 g/(0.879 g/mL)

V = 150.1 mL

Explanation:

At convergent plate boundaries two plates are moving towards one another and different scenarios are in play.

Most importantly, at these margins the crust is being destroyed.

Some of the features found at this margin include, oceanic trenches, accretionary wedges, volcanic islands, earthquakes

Answer: ice is less dense than liquid water. If ice was more dense, Earth would freeze.

Explanation: There are many reasons why life on Earth depends on the characteristics of water. One could discuss hydrogen bonds and its role as a solvent, but the unusual property of water is is the change in density with change in temperature. Water is densest at 4 degC, which is why ice floats - it is less dense than cold water (it melts quickly in warm water, so density isn’t impotant at higher temperatures). Most liquids are less dense than the solid, frozen form. If this was the case with water, any ice that formed would sink, and sease would freeze from the bottom up. Furthermore, the lowest layers would be insulated and would not all melt in summer. Thus over time, the seas would become a thin layer of liquid water at best, over solid ice. Life could not develop without liquid seas. In addition, ice is reflective, reducing the amount of sunlight absorbed, further reducing temperatures. Without ocean circulation, polar areas would be even colder, and there would be no rain.

Answer:

Cccccccccccccccccccccccccccccccccccccccccccc

Answer:

[tex]pH=4.63[/tex]

Explanation:

Hello!

In this case, since the ionization of ammonia, which is a weak base, is written as:

[tex]NH_3+H_2O\rightleftharpoons NH_4^++OH^-[/tex]

We can see that the ammonium ion is the conjugate acid whereas the hydroxide ions the conjugate base; that is why we use the Henderson-Hasselbach equation to compute the pH, given the pKb of ammonia 4.75:

[tex]pH=pKb+log(\frac{[conj\ acid]}{[base]} )[/tex]

In such a way, for the given moles of ammonia, base, and those of ammonium chloride, conjugate acid form, we obtain:

[tex]pH=4.75+log(\frac{0.15}{0.2} )\\\\pH=4.63[/tex]

Best regards!

Answer:

B. a chemical reactions that occurs in the chloroplast of a plant.

Explanation:

Answer:

B.

Explanation:

it is b because I took science

lol have a good day

Answer: B

Mutualism.

Explanation:

Answer:

The answer is "35 atoms".

Explanation:

The Formula for isopropyl alcohol is: [tex]C_3H_8O[/tex]

In 1 molecule of [tex]C_3H_8O[/tex] = 8 atoms of hydrogen  

In 4 molecule of = atoms of hydrogen

According to the given formula [tex]C_3H_7O[/tex]

In 5 molecule of [tex]C_3H_7O[/tex] [tex]= 5 \times 7 = 35[/tex] atoms of hydrogen.

Answer:

The answer is C or 3 which ever but it that one

Explanation:

Answer:

I don't think that answer is right. I got 543.04 grams.

Explanation:

I think the question is asking for grams in the samlpe instead of the moles in the sample because it already gives you the amount of moles.

Answer:

[tex]r=0.0341\frac{M}{s}[/tex]

Explanation:

Hello!

In this case, considering the given table, we are able to represent the symbolic rate law as shown below:

[tex]r=k[BF_3]^m[NH_3]^n[/tex]

Thus, by using the following steps, we can find both m (BF3 order of reaction) and n (BF3 order of reaction):

- Experiment 1 and 2 for the calculation of n:

[tex]\frac{r_1}{r_2}=\frac{k[BF_3]_1^m[NH_3]_1^n}{k[BF_3]_2^m[NH_3]_2^n}[/tex]

So we plug in to obtain:

[tex]\frac{0.2130}{0.1065}=\frac{k*0.250^m*0.250^n}{k*0.250^m*0.125^n}\\\\2=\frac{0.250^n}{0.125^n}\\\\2=2^n\\\\log(2)=n*log(2)\\\\n=1[/tex]

So the order of reaction with respect to NH3 is 1.

- Experiment 3 and 4 for the calculation of m

[tex]\frac{r_3}{r_4}=\frac{k[BF_3]_3^m[NH_3]_3^n}{k[BF_3]_4^m[NH_3]_4^n}[/tex]

So we plug in to obtain:

[tex]\frac{0.0682}{0.1193}=\frac{k*0.200^m*0.100^n}{k*0.350^m*0.100^n}\\\\0.57=0.57^m\\\\log(0.57)=m*log(0.57)\\\\m=1[/tex]

So the order of reaction with respect to BF3 is also 1.

Now, we can compute the rate constant by solving for it on any of the experiments there, say experiment 1:

[tex]k=\frac{r_1}{[BF_3][NH_3]} =\frac{0.2130M/s}{0.250M*0.250M}\\\\k=3.41M^{-1}s^{-1}[/tex]

Thus, the initial reaction rate for the 0.50M BF3 and 0.020M NH3 is:

[tex]r=3.41M^{-1}s^{-1}*0.50M*0.020M\\\\r=0.0341\frac{M}{s}[/tex]

Best regards!

Answer:

[tex]\mathbf{H_2SeO_{4(aq)} + 3SO_{2(aq)} + 2H_2O_{(l) } \to Se_{(s)} + 3H_2SO_{4(aq)} }[/tex]

Explanation:

We are to write a balanced equation from the information given above by including the states of their matter according to SATP conditions;

[tex]\mathbf{H_2SeO_{4(aq)} + 3SO_{2(aq)} + 2H_2O_{(l) } \to Se_{(s)} + 3H_2SO_{4(aq)} }[/tex]

From above;

The oxidizing agent is [tex]\mathbf{H_2SeO_{4(aq)} }[/tex]

The reducing agent is [tex]\mathbf{SO_{2(aq)} }[/tex]

Answer: When the spring is recoiling

Explanation:

Answer:

When the spring is recoiling

Explanation:

Answer:

This region of thick, hot crust and mantle extends 1,300 kilometers north and south of the Iceland hotspot and some 1,000 kilometers north and south of the Azores hotspot. Together, the two hotspots appear to be feeding a huge supply of magma to nearly the entire northern segment of the Mid-Atlantic Ridge.

Explanation:

please give me a heart

Answer:

Easter island

Explanation:

I mol of water = 22.4 lts of O2

thus, 4.21 moles of water = 4.21× 22.4= 94.304 lts.

Answer:

See explanation

Explanation:

An initiation step is the first step in a free radical mechanism. In this step, free radicals are generated from a stable compound. The initiation step tends to increase the concentration of radicals present in the system.  It leads to a net increase in the number of free radicals in the system.

For the chlorination of 2,2-dimethylpropane, the initiation step is;

Cl2 + hv --------->2Cl.

Note that hv is used to represent light

Answer:

Explanation:

Airs and trees

Answer:

no they do not

Explanation:

Answer:

It's just...Hair grown there...i guess it just grows as a male humans would

Explanation:

hope this helps.

Answer:

A control is important for an experiment because it allows the experiment to minimize the changes in all other variables except the one being tested.

Explanation:

do u mean that kind of control

The given question is incomplete. The complete question is:

A chemist adds 0.85 L of a 0.0050M calcium sulfate to a reaction flask. Calculate the mass in grams of calcium sulfate the chemist has added to the flask. Round your answer to significant digits.

Answer: The mass in grams of calcium sulfate the chemist has added to the flask is 0.58 g

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n}{V_s}[/tex]

where,

n = moles of solute

[tex]V_s[/tex] = volume of solution in L

moles of [tex]CaSO_4[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{136g/mol}[/tex]

Now put all the given values in the formula of molarity, we get

[tex]0.0050=\frac{x}{136\times 0.85}[/tex]

[tex]x=0.58 g[/tex]

Therefore, the mass in grams of calcium sulfate the chemist has added to the flask is 0.58 g

4 cm³

Math

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Chemistry

Gas Laws

Density = Mass over Volume

Step 1: Define

D = 25 g/cm³

m = 100 g

Step 2: Solve for V

Answer:

Hate 2 break it 2 u foo but there's no option for me 2 choose....

290 g Na₂O

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

4.7 mol Na₂O

Step 2: Identify Conversions

Molar Mass of Na - 22.99 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Na₂O - 2(22.99) + 16.00 = 61.98 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

291.306 g Na₂O ≈ 290 g Na₂O