1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Helen [10]
4 years ago
15

PROJECT: KITCHEN TOOLS

Chemistry
1 answer:
docker41 [41]4 years ago
7 0

Answer:

Explanation:2+2=4

You might be interested in
Describe what happens to the molecules in the ice as temperature of the ice increases
german

Answer:

When you heat ice, the individual molecules gain kinetic energy, but until the temperature reaches the melting point, they don't have energy to break the bonds that hold them in a crystal structure. They vibrate more quickly within their confines as you add heat, and the temperature of the ice goes up.

8 0
3 years ago
Read 2 more answers
How many moles of a gas sample are in a 10.0 L container at 298 K and 203 kPa? (Use the ideal gas law PV = nRT with R = 8.31 L-k
weeeeeb [17]
Ideal gas law: PV = nRT

P-pressure
V-volume
n-number of moles(m/M)
R-constant
T- temperature

State given info:
V=10L
T=298k
P=203kPa
R=8.31(only for kPa, for pressure at atm use 0.08206)

Sub in:
(203kPa)(10L)=n(8.31)(298)

Rearrange:
n= (203)(10) / (8.31)(298)
n = 0.819745mol in gas sample
8 0
3 years ago
Which compound becomes less soluble in water as the temperature of the solution is increased?(1) HCl (2) KCl (3) NaCl (4) NH4Cl
ratelena [41]
HCI are less soluble in water as the temperature of the solution is increased.
3 0
4 years ago
Read 2 more answers
Calculate the cost of 1.00 g of gold if the price gold is 655.00 per oz
Anna007 [38]

Answer:

22.96 $ per 1 g of gold

Explanation:

We know that 1 g is equal to 0.035 oz.

Now we formulate the following reasoning:

if         1 oz of gold have a value of 655 $

then    0.035 oz of gold have a value of X $

X = (0.035 × 655) / 1 = 22.96 $

3 0
3 years ago
Read 2 more answers
How many grams of sodium formate, NaCHO2, would have to be dissolved in 3.0 L of 0.12 M formic acid (pKa 3.74) to make the solut
kolbaska11 [484]

Answer:

Mass_{sodium\ formate}= 889.57\ g

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the acidic buffer solution as:

pH=pK_a+log\frac{[salt]}{[acid]}

Given that:-

[Acid] = 0.12 M

Volume = 3.0 L

pKa = 3.74

pH = 5.30

So,  

5.30=3.74+log\frac{[sodium\ formate]}{0.12}

Solving, we get that:-

[Sodium formate] = 4.36 M

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,

Moles =Molarity \times {Volume\ of\ the\ solution}

So, Moles of sodium formate = 4.36*3.0 moles = 13.08 moles

Molar mass of sodium formate = 68.01 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

13.08\ mole= \frac{Mass}{68.01\ g/mol}

Mass_{sodium\ formate}= 889.57\ g

5 0
4 years ago
Other questions:
  • Use the periodic table to give the name and symbol for each of the following elements:
    12·1 answer
  • Identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent for the following equation
    6·1 answer
  • How many is 1 min = ? km
    12·1 answer
  • A patient receives an injection of 2.0 ml of anesthetic. what is the equivalent of this volume in cubic centimeters?
    7·1 answer
  • Calculate Delta E when 33.0 g of carbon dioxide sublimes at 77.0 K and 1 ATM
    15·1 answer
  • How can i find nuclear charge and overall charge while only knowing the atomic mumber, number of protons and electrons?​
    9·1 answer
  • Here some points to you peeps.​
    8·2 answers
  • 7. Which structural formula correctly represents a hydrocarbon molecule?
    11·1 answer
  • Need help with this can someone help me pls the word box are words your supposed to use. Will mark brainiest btw
    14·1 answer
  • Chất nào sau đây là chất điện li mạnh trong H2O
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!