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tatuchka [14]
3 years ago
9

The image above shows a chamber with a fixed volume

Chemistry
1 answer:
kogti [31]3 years ago
8 0

Answer: 1094

Explanation:

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If the pattern shown continued, the next low tide occurred on Day 3 at approximately
romanna [79]
2) 1 :30 pm 


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8 0
3 years ago
Read 2 more answers
Vanadium (IV) carbonate
Anna71 [15]

Answer:

Explanation:

1) Vanadium (IV) → V⁺⁴

Carbonate → CO₃⁻²

So , Vanadium (IV) Carbonate = V₂(CO₃)₄  or V(CO₃)₂

2) Tin (II) = Sn⁺²

Nitrite = NO₂⁻

So, Tin (II) Nitrate = Sn(NO₂)₂

3) Cobalt (III) = Co⁺³

Oxide = O⁻²

So , Cobalt (III) Oxide = Co₂O₃

4) Titanium (II) = Tn⁺²

Acetate = CH₃COO⁻

So , Titanium (II) Acetate = Tn(CH₃COO)₂ or Tn(C₂H₃O₂)₂

5) Vanadium (V) = V⁺⁵

Sulfide = S⁻²

So , Vanadium (V) Sulfide = V₂S₅

6) Chromium (III) = Cr⁺³

Hydroxide = OH⁻

So , Chromium (III) Hydroxide = Cr(OH)₃

7) Lithium = Li⁺

Iodide = I⁻

So , Lithium Iodide = LiI

8) Lead (II) = Pb⁺²

Nitride = N⁻³

So , Lead (II) Nitride = Pb₃N₂

9) Silver = Ag⁺

Bromide = Br⁻

So , Silver Bromide = AgBr

5 0
3 years ago
Suppose a substance has a heat of fusion equal to 45 calg and a specific heat of 0.75
Varvara68 [4.7K]

Answer:

The substance will be in liquid state at a temperature of 97.3 °C

Note: The question is incomplete. The complete question is given below :

Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)

Explanation:

1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT

Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C

H = 50 × 0.75 × 3 = 112.5 calories

b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g

H = 50 × 45 = 2250 cal

c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.

Amount of energy left = 5000 - 2362.5 = 2637.5 cal

The remaining energy is used to heat the liquid

H = mcΔT

Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change

2637.5 = 50 × 0.75 x ΔT

ΔT = 2637.5 / ( 50*0.75)

ΔT = 70.3 °C

Final temperature of sample = (70.3 + 27) °C = 97.3 °C

The substance will be in liquid state at a temperature of 97.3 °C

5 0
3 years ago
__________ ammonium compounds are less effective against gram-negative bacteria and more effective against gram-positive bacteri
Marina86 [1]

Answer:

Quartenary.

Explanation:

Quaternary ammonium compounds (QACs) are cationic salts of organically substituted ammonium compounds and have a broad range of activity against microorganisms, i.e more effective against Gram-positive bacterium at lower concentrations than Gram-negative bacteria.

It was also reported previously that monoalkyl QACs bind by ionic and hydrophobic interactions to microbial membrane surfaces, with the cationic head group facing outwards and the hydrophobic tails inserted into the lipid bilayer, causing the rearrangement of the membrane and the subsequent leakage of intracellular constituents.

6 0
3 years ago
Explain how you would calculate the q for warming 100.0 grams of liquid water from 0°C to 100 °C.
mojhsa [17]

Answer:

mass = 100 g

T1 = 0°C

T2 = 100 °C

C = 1 cal/g°C

Q = mC(T2 -T1)

Q = 100(1)(100 - 0)

Q = 100(100)

Q = 10000 cal

Explanation:

6 0
3 years ago
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