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BartSMP [9]
3 years ago
9

An object has a mass of 35.0 grams.  On Anthony’s balance, it weighs 34.85 grams.  What is the percent error of his balance?

Chemistry
1 answer:
blondinia [14]3 years ago
5 0
34.85 is his balance
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Will give brainlsit need answers asap!
k0ka [10]

Answer:

120 g of NaCl in 300 g H20   at 90 C

Explanation:

At x = 90   go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20   = 40

we want   300 g H20    so multiply this by  3    to get   120 gm of NaCl in 300 g

7 0
2 years ago
Which of these is NOT a property of water?
Elis [28]

Answer:

-low specific heat

3 0
3 years ago
Read 2 more answers
Be<br><br> Name the element.<br><br> Number of shells?<br><br> Valence electrons?
Vedmedyk [2.9K]

Answer:

Name the element: Beryllium

Number of shells:  4

Valence electrons: 2

Explanation:

4 0
3 years ago
What process forms an image in a mirror? Choices: A. absorbing light B. Reflecting light C. Refracting light D. Transmitting lig
Dafna1 [17]

Answer

Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, since it cannot be projected—the rays only appear to originate from a common point behind the mirror.

Explanation:

Hope this helps someone

4 0
3 years ago
Consider the following balanced equation:
algol [13]

Moles of PF₃ : 4

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

\tt P_4(s)+6F_2(g)\rightarrow 4PF_3(g)

1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)

Limiting reactant : the smallest ratio (mol divide by coefficient)

P₄ : F₂ =

\tt \dfrac{1.25}{1}\div \dfrac{6}{6}=1.25\div 1\rightarrow F_2~limiting~reactant(smallest~ratio)

mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :

\tt \dfrac{4}{6}\times 6~moles=4~moles

8 0
3 years ago
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