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Margarita [4]
2 years ago
5

LAB: predicting products

Chemistry
1 answer:
kondor19780726 [428]2 years ago
4 0

Answer:

1) synthesis MgI2

2) double replacement CuS + (HCl)2

3) double replacement, not sure ab the formula sorry

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Anil accidentally swallows a substance that activates the complement cascade in the absence of bound antibodies. this substance
ale4655 [162]

Answer:

This substance would cause cause an extensive cellular damage to Anil.

5 0
1 year ago
A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

  • E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ} and
  • m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}

Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
  • Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}
  • Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
  • E(\text{released}) =  Q(\text{lead})

Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0
3 years ago
How many grams of aluminum chloride must decompose to produce 78.3 milliliters of aluminum metal, if the density of aluminum is
Vilka [71]
1mol aluminium chloride gives 1mol aluminium and 3mol chloride
density equals mass divided by volume
d=m/v
m=v*d
=78.3*2.7
=211.41grams
7 0
3 years ago
What are the classes of alkanol
LuckyWell [14K]
1-PRIMARY ALKANOL 2-SECONDARY ALKANOL 3-TERTIARY ALKANOL
8 0
3 years ago
Calculate the volume of 1M NaOH required to neutralize 200 cc of 2M HCI. What mass of sodium
Nezavi [6.7K]

NaOH+HCl-> NaCl+H2O

1 mole of NaOH

1 mole of HCl.

To calculate volume of NaOH

CaVa/CbVb= Na/Nb

Where Ca=2M

Cb=1M

Va=200cm³

Vb=xcm³

Substitute into the equation.

2×200/1×Vb=1/1

400/Vb=1/1

Cross multiply

Vb×1=400×1

Vb=400cm³

To calculate the mass of sodium chloride, NaCl from the neutralization rxn.

Mole of NaCl=1

Molar mass of NaCl= 23+35.5=58.5

Mass=xgrammes.

Mass of NaCl=Number of moles × Molar mass.

Substitute

Mass of NaCl= 1×58.5

=58.5g

This is what I could come up with.

5 0
3 years ago
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