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Allisa [31]
2 years ago
11

CCan you refrigerate breast milk that has been warmed?.

Chemistry
1 answer:
Nookie1986 [14]2 years ago
3 0
Yes if you don’t believe me just search it up
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Please help quick the 6th question is “What is a catalyst?” (include all three parts)
allochka39001 [22]
1. The reaction is exothermic.

2. I know this because the delta H is a negative number. (-802.4 kJ)

3. This means that this reaction releases heat.
4 0
3 years ago
1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.
Snowcat [4.5K]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of CH_3OH = 26.5 g

Molar mass of CH_3OH = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{26.5\ g}{32.04\ g/mol}

Moles\ of\ CH_3OH= 0.8271\ moles

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

m=\frac {0.8271\ moles}{0.735\ kg}

<u>Molality = 1.13 m</u>

4 0
3 years ago
Energy comes in different forms.
LenKa [72]

Answer:

liquid, solid, and gas

Explanation: It depends where the molecules are moving. When a solid the molecules are vibrating and are all together, compact, the molecules are also very slow. When a liquid the molecules are moving back and forth, up and down, and are less compact, but moving faster. When a gas, the molecules move everywhere very quickly, moving super fast.

3 0
3 years ago
Which is more concentrated, a solution containing 10g of NaCl in 100 g
Pavel [41]

Answer:

The answer is the solution containing the 90g of water.

Explanation:

Because both solutions have the same amount of the sodium chloride, the concentration would be the same if they both had the same amount of water.

Because the second solution has less water, there is more of the sodium chloride per each drop of water.

I hope this helps :)

4 0
3 years ago
HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into
dsp73

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.037 M

<u>Explanation:</u>

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

                       2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>               1.0

<u>At eqllm:</u>        1.0-2x          x           x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

Kc=1.6\times 10^{-3}

Putting values in above expression, we get:

1.6\times 10^{-3}=\frac{x\times x}{(1.0-2x)^2}\\\\x=-0.043,0.037

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.037 M

Hence, the concentration of hydrogen gas at equilibrium is 0.037 M

6 0
3 years ago
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