All categories

2 years ago

Chemistry homework about elements and compounds.

Chemistry

1 answer:

galben [10]2 years ago

8 0

1. 3 elements, Mg: 1, S: 1, O: 4
2. 3 elements, Li: 3, P: 1, O: 4
3. 3 elements, H: 2, S: 1, O: 4

1. KOH
2. AlOH
3. AlSO4

You might be interested in

Not sure what to do can you help

NeTakaya

sorry i don't know the answer i'm really sorry

7 0

2 years ago

How many electrons must Aluminum lose or gain to obtain a noble gas electron configuration? *

BaLLatris [955]

Answer:

3 electrons

Explanation:

aluminum : [Ne]3s23p1 [ N e ] 3 s 2 3 p 1 . It loses 3 electrons from 3s and 3p orbitals and attains the noble gas configuration of Neon.

6 0

3 years ago

Read 2 more answers

A 40.0 L balloon is filled with air at sea level (1 atm @ 25 oC). It is then tied to a rock and thrown into a cold lake and it s

velikii [3]

To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-1 * 298.15K)
n= 1.636 moles

The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= 3.38 L

8 0

3 years ago

2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

4 0

3 years ago

Complete the table below by deciding whether a precipitate forms when aqueous solutions A and B are mixed. If a precipitate will

stiks02 [169]

The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.

The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.

Now, we will consider each reaction individually to decode whether or not a precipitate is possible.

In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.

Learn more: brainly.com/question/11969651

6 0

3 years ago