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Bumek [7]
2 years ago
11

A 141mg sample was placed on a watch glass that has a mass of 9.203g. what is the mass of the watch glass and sample in grams?

Chemistry
1 answer:
Volgvan2 years ago
8 0

Given the mass of sample placed on the watch glass = 141 mg

Converting mass of the sample from mg to g:

141 mg * \frac{1 g}{1000 mg} =0.141 g

Mass of watch glass = 9.203 g

So, the total mass of watch glass and the sample = 9.203 g +0.141 g = 9.344 g

Therefore, the mass of the watch glass and the sample in grams is 9.344 g

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What volume of a 0.716 m kbr solution is needed to provide 30.5 g of kbr?
Jlenok [28]
Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
7 0
1 year ago
Read 2 more answers
How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made
natta225 [31]
Answer:  
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.  
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn  
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced  
Now use the ideal gas law to find the volume V.  
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm  
T = 21 C + 273 = 294 K  
PV = nRT 
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L
7 0
2 years ago
Two students titrated a 25.0 mL aliquot of pear juice with 0.107 M NaOH to the phenolphthalein end point. The initial buret read
Olin [163]

Answer:

0.363g citric acid

Explanation:

Sodium hydroxide (NaOH) reacts with acids, thus:

NaOH + H⁺ → H₂O + Na⁺

The volume of titration is:

18.39mL - 0.73mL = 17.66mL

Moles of this volume in 0.107M NaOH are:

0.01766L × (0.107 mol / L) = 0.00189mol NaOH ≡ mol citric acid<em> -Assuming the only acid in pear juice is citric acid-</em>

As molar mass of citric acid is 192.124g/mol:, Mass of citric acid is:

0.00189mol citric acid × (192.124g / mol) = <em>0.363g citric acid</em>

I hope it helps!

6 0
2 years ago
What is water's density at 93 ∘C? Assume a constant coefficient of volume expansion. Express your answer with the appropriate un
Ganezh [65]

Answer:

982.5 kg/m³

Explanation:

When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:

ρ₁ = ρ₀/(1 + β*(t₁ - t₀))

Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.

At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C

ρ₁ = 1,000/(1 + 0.0002*(93 - 4))

ρ₁ = 1,000/(1+ 0.0178)

ρ₁ = 982.5 kg/m³

3 0
2 years ago
How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K
Olenka [21]

I’m not sure I need help with this question

6 0
1 year ago
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