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3 years ago

When exercising what happens to the heart rate and blood movement?

Chemistry

1 answer:

Slav-nsk [51]3 years ago

8 0

Your heart rate increases and your blood pressure increases, causing blood to flow around your body faster.<span />

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Count the total number of atoms in sio2:

grigory [225]

Atoms are the basic units of matter and the defining structure of elements. We count the number of atoms by the total number of elements present in the compound. In this case, we have 1 atom of Si and 2 atoms of oxygen which would have 3 total number of atoms.

8 0

3 years ago

HELP!!!! ASAP!!!!!!!!!!<br><br> HOW ARE REACTANTS DIFFERENT FROM THE PRODUCTS ?

raketka [301]

Reactants are what make up the product.

7 0

3 years ago

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

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Which type of bond is MOST LIKELY formed in nitrous oxide?

arsen [322]

The type of bond most likely formed in nitrous oxide would be a covalent bond..

8 0

3 years ago

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If the volume occupied by 0.500 mol of nitrogen gas at 0°C is 11.2 L, then the volume occupied by 2.00 mol of nitrogen gas at th

Paraphin [41]

Avagadros law states that volume of gas is directly proportional to number of moles of gas at constant pressure and temperature.
$\frac{V1}{n1} = \frac{V2}{n2}$
where V -volume , n - number of moles
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
$\frac{11.2 L}{0.500 mol} = \frac{V}{2.00 mol}$
V = 44.8 L
answer is C. 44.8 L

5 0

3 years ago