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3 years ago

- Name 2 scientists that were the first to observe cells.

Chemistry

2 answers:

ExtremeBDS [4]3 years ago

3 0

one was Robert Hooke but idk the other one

crimeas [40]3 years ago

3 0

Answer: Theodor Schwann and Matthias Jakob Schleiden

Explanation:

Theodor Schwann and Matthias Jakob Shleiden are German scientists that were the first to observe cells.

Around 1850, a German doctor named Rudolf Virchow was studying cells under a microscope, when he happened to see them dividing and forming new cells.

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7. A student has a 25 g sample of gas in a container at STP and adds another 7

BigorU [14]

Answer:

New temperature t2 = [1.28T− 273.15]° C

Explanation:

Given:

Volume v1 = 25 gram

New volume v2 = 25 + 7 = 32 gram

Constant pressure = p

Temperature t1 = T

Find:

New temperature t2

Computation:

Pv1/t1 = Pv2/t2

25 / T = 32 / t2

t2 = 1.28T

New temperature t2 = [1.28T− 273.15]° C

4 0

3 years ago

Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0

3 years ago

What is the interval used on the y-axis of this graph?

bezimeni [28]

Answer:

Explanation:

the numbers go up by 5

7 0

3 years ago

If element x forms the oxides xo and x2o3 the oxidation numbers of element x are

Deffense [45]

Answer:

b) +2 and +3.

Explanation:

Hello,

In this case, given the molecular formulas:

$XO$

And:

$X_2O_3$

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

$X^{+2}O^{-2}$

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

$X_2^{+3}O_3^{-2}$

Thus, answer is b)+2 and +3

Best regards.

3 0

3 years ago

How many grams of CO2 could be formed with 2.09 x 1023 atoms of O?

choli [55]

Answer:

r u in high school this is hard

Explanation:

6 0

3 years ago