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Alexxx [7]
3 years ago
5

What could cause a test tube to crack?

Chemistry
2 answers:
ladessa [460]3 years ago
7 0
Heat they’re not meant to be used with hot chemicals
Molodets [167]3 years ago
5 0

Answer:

heated test tube in cold water

You might be interested in
3 points
VLD [36.1K]

Answer:

<h3>The answer is 8.29 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

p(\%) =  \frac{1.6}{19.3}  \times 100 \\  = 8.290155440...

We have the final answer as

<h3>8.29 %</h3>

Hope this helps you

5 0
3 years ago
What are the products in a chemical reaction?
zlopas [31]
Products are the species formed from chemical reactions. During a chemical reaction reactants are transformed into products after passing through a high energy transition state. This process results in the consumption of the reactants.
7 0
2 years ago
The fifth element in group 7 in astatine would you expect it to be gas, liquid, or solid? why?
Ymorist [56]

Answer:

solid

Explanation:

Melting and boiling points of Group 7 elements State at room temperature Room temperature is usually taken as being 25°C. At this temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine and astatine are solids. There is therefore a trend in state from gas to liquid to solid as you go down the group.

6 0
2 years ago
3. How many formula units of NaOH are in 0.87 moles?
Lady_Fox [76]

Answer: 1 mole of H2O= about 1/3 of a cup (18 mL). It is helpful ... 6.02 x 1023 H2O molecules. = 6.02 x 1023 NaCl formula unit. 1 mole C. 1 mole H2O. 1 mole

Explanation:

5 0
2 years ago
Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai
Zarrin [17]

Answer : The concentration of HI and I_2 at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of H_2, I_2\text{ and }HI

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M

\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M

\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                            2HI(g)\rightleftharpoons H_2(g)+I_2(g)

Initial conc.      0.0163     0.00415      0.00276

At eqm.        (0.0163-2x) (0.00415+x)  (0.00276+x)

As we are given:

Concentration of H_2 at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of HI at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of I_2 at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

8 0
3 years ago
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