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Fofino [41]
4 years ago
11

Grade

Chemistry
1 answer:
Anuta_ua [19.1K]4 years ago
4 0

Answer:

pp

Explanation:

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I need help as fast as possible
Kaylis [27]

Answer:

its c i did it not tolong ago

Explanation:

6 0
3 years ago
Classify these salts as acidic, basic, or neutral. KCL, NH4Br, K2CO3, NaCN, LiClO4? This is what I put but got it wrong: KCl = b
MaRussiya [10]
In classifying salts as neutral, acidic, or basic, it is important to take note of the strength of the acids and bases that they come from. A strong acid and strong base produce a neutral salt. A weak acid and strong base produce a basic salt. A strong acid and weak base produce an acidic salt. So the answers must be:

KCl = neutral (from HCl and KOH)
NH4Br = acidic (from NH4 and HBr)
K2CO3 = basic (from KOH and H2CO3)
NaCN = basic (from NaOH and HCN)
LiClO = basic (from LiOH and HClO)
3 0
4 years ago
Read 2 more answers
A gas occupies 3.8 L at -18° C and 975. torr. What volume would this gas occupy at STP?
Aleks04 [339]

To solve the question we will assume that the gas behaves like an ideal gas, that is to say, that there is no interaction between the molecules. Assuming ideal gas we can apply the following equation:

PV=nRT

Where,

P is the pressure of the gas

V is the volume of the gas

n is the number of moles

R is a constant

T is the temperature

Now, we have two states, an initial state, and a final state. The conditions for each state will be.

Initial state (1)

P1=975Torr=1.28atm

V1=3.8L

T1=-18°C=255.15K

Final state(2), STP conditions

P2=1atm

T2=273.15K

V2=?

We will assume that the number of moles remains constant, so the nR term of the first equation will be constant. For each state, we will have:

\begin{gathered} \frac{P_1V_1}{T_1}=nR \\ \frac{P_2V_2}{T_2}=nR \end{gathered}

Since nR is the same for both states, we can equate the equations and solve for V2:

\begin{gathered} \frac{P_{2}V_{2}}{T_{2}}=\frac{P_1V_1}{T_1} \\ V_2=\frac{P_{1}V_{1}}{T_{1}}\times\frac{T_2}{P_2} \end{gathered}

We replace the known values:

V_2=\frac{1.28atm\times3.8L}{255.15K}\times\frac{273.15K}{1atm}=5.2L

At STP conditions the gas would occupy 5.2L. First option

8 0
1 year ago
Why can't all mixtures be classified as solutions?
shusha [124]
D. Not all mixtures are heterogeneous
7 0
3 years ago
How many grams of potassium chloride will be needed to produce<br> 829 grams of zinc chloride?
grandymaker [24]

Answer:

[tex]2KCl + Zn {}^{2 + } → 2K {}^{ + } + ZnCl _{2} \\ molecular \: mass \: of \: zinc \: chloride = 65 + (35.5 \times 2) = 136 \: g \\ molecular \: mass \: of \: potassium \: chloride = 39 + 35.5 = 74.5 \: g

6 0
3 years ago
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